Solving Quadratic Equations

This deck covers every way to solve a quadratic equation, and how to pick the fastest one. It starts with standard form and the zero-product property, then works through factoring, the square root property, completing the square, the quadratic formula derived from it, and what the discriminant says about the number and type of solutions, before finishing with projectile, area, revenue, and Pythagorean applications. It targets the four errors that cost the most points: setting factors equal to a constant instead of zero, losing half the solutions by dropping the plus-or-minus, dropping the sign of a negative coefficient inside the quadratic formula, and dividing only one term by the leading coefficient while completing the square.

Subject: College Algebra · 146 slides · symbolic lesson

Open the interactive version of this deck · Homework for this lesson

What this lesson covers

The lesson, slide by slide

1. Solving Quadratic Equations

Title

College Algebra - Deck 06

Four methods, one decision rule for choosing between them, and the discriminant that tells you what you are about to find.

2. What you will be able to do

Objectives

A linear equation has one answer. A quadratic usually has two - and most lost points in this unit come from finding only one of them.

  1. Put any quadratic equation into standard form and read off its three coefficients with their signs.
  2. Solve by factoring using the zero-product property, and say why the equation has to equal zero first.
  3. Solve by the square root property without ever losing the negative half of the answer.
  1. Complete the square, including when the leading coefficient is not one.
  2. Use the quadratic formula on any quadratic, substituting negative coefficients safely.
  3. Read the discriminant and predict the number and type of solutions before solving.
  4. Set up and solve applied quadratics - height, area, revenue, right triangles - and reject the answers the context forbids.

3. What survived from Absolute Value Equations and Inequalities?

Warm-up

Discussion prompt

Before we open Solving Quadratic Equations: without looking back, what was the main idea of Absolute Value Equations and Inequalities, and what could you do by the end of it that you could not do before?

Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.

Answer:

That deck teaches absolute value from a single picture - distance from zero on the number line - and then uses it to solve every standard problem type: isolate-then-split equations, the no-solution and single-solution cases, absolute value on both sides, less-than inequalities as one interval, greater-than inequalities as a union of two rays, the always-true and never-true cases, and tolerance and error-bound applications. It targets the four errors that sink students here: splitting before isolating, swapping "and" with "or", negating only the constant on the other side, and forgetting to check candidates against a variable right-hand side.

4. What Makes an Equation Quadratic

Section

Part 1

5. Standard form is the starting line

Concept

An equation is quadratic when you can rearrange it so one side is zero and the other side has a squared variable as its highest power.

\[ ax^2 + bx + c = 0 \]

standard form — A quadratic written with every term on one side, in descending powers, and zero alone on the other side. The three coefficients a, b, and c are read off directly from it - signs included.

6. Break it if you can: Standard form is the starting line

Counterexample

Discussion prompt

An equation is quadratic when you can rearrange it so one side is zero and the other side has a squared variable as its highest power.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

7. Vertex form shows the vertex

Picture it

Animation

Shows: Vertex form shows the vertex — a rendered Manim animation.

Rendered with Manim.

Takeaway: The number inside the bracket moves the parabola horizontally.

8. The leading coefficient cannot be zero

Concept

The squared term is what makes it quadratic. If its coefficient is zero, the squared term disappears and you are back to a linear equation.

\[ a \ne 0 \]

The other two coefficients are allowed to be zero. Both of these are still quadratic equations, just with missing pieces:

\[ x^2 - 9 = 0 \qquad\text{and}\qquad 3x^2 + 12x = 0 \]

9. By analogy: The leading coefficient cannot be zero

Analogy

Discussion prompt

Explain The leading coefficient cannot be zero by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Answer:

The squared term is what makes it quadratic. If its coefficient is zero, the squared term disappears and you are back to a linear equation.

10. Picture it first: Why a quadratic usually has two answers

Picture it

Figure (svg): A U-shaped parabola crossing a horizontal axis at two marked points

A curve that dips below an axis has to cross it twice.

Discussion prompt

Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.

Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.

Answer:

Graph the left side and you get a parabola - a U-shaped curve. Solving the equation means asking where that curve sits at height zero.

11. Why a quadratic usually has two answers

Intuition

Graph the left side and you get a parabola - a U-shaped curve. Solving the equation means asking where that curve sits at height zero.

Figure (svg): A U-shaped parabola crossing a horizontal axis at two marked points

A curve that dips below an axis has to cross it twice.

A curve that comes down, crosses, and goes back up crosses twice. So when you finish a quadratic with one answer, that is a claim you should double-check, not a normal result.

12. Teach it back: Why a quadratic usually has two answers

Explain it

Discussion prompt

Explain Why a quadratic usually has two answers to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Answer:

Graph the left side and you get a parabola - a U-shaped curve. Solving the equation means asking where that curve sits at height zero.

13. Every method starts the same way

Concept

Factoring, the quadratic formula, and the discriminant all read the coefficients off standard form. Rearranging first is not busywork - it is the step that makes the rest legal.

14. What has to happen first: Worked example: rewrite in standard form

Ranking

Put in order

Put the moves of Worked example: rewrite in standard form into the order they have to happen.

  1. Distribute on the left
  2. Subtract 5x and subtract 4 from both sides
  3. Read the coefficients with their signs
  4. Verify by undoing the rearrangement

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Nothing can be collected while a term is trapped inside parentheses.

15. Worked example: rewrite in standard form

Worked example

Put this equation in standard form and name its three coefficients.

\[ 3x(x - 2) = 5x + 4 \]

Distribute on the left

Why: Nothing can be collected while a term is trapped inside parentheses.

\[ 3x^2 - 6x = 5x + 4 \]

Subtract 5x and subtract 4 from both sides

Why: Move every term to the left so the right side becomes zero; the squared term is already positive there.

\[ 3x^2 - 11x - 4 = 0 \]

Read the coefficients with their signs

Why: The minus signs belong to the numbers. Carrying them now prevents the most common formula error later.

\[ a = 3, \quad b = -11, \quad c = -4 \]

Verify by undoing the rearrangement

Why: Add 5x and 4 back to both sides of the standard form: it returns to 3x squared minus 6x equals 5x plus 4, and the left side factors back to 3x times the quantity x minus 2 - the original equation.

\[ 3x^2 - 11x - 4 = 0 \;\Longleftrightarrow\; 3x^2 - 6x = 5x + 4 \;\Longleftrightarrow\; 3x(x-2) = 5x + 4 \]

16. rewrite in standard form — line by line

Picture it

Animation

Shows: Each line of the worked example "rewrite in standard form", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Add 5x and 4 back to both sides of the standard form: it returns to 3x squared minus 6x equals 5x plus 4, and the left side factors back to 3x times the quantity x minus 2 - the original equation.

17. Rebuild the recipe: Pattern: get to standard form

Ranking

Put in order

These are the steps of Pattern: get to standard form, scrambled. Put them back in order before the next slide shows you.

  1. Clear parentheses by distributing, and clear fractions by multiplying every term by the common denominator.
  2. Move all terms to one side so the other side is exactly zero.
  3. Choose the side that leaves the squared term positive - it makes factoring and the formula easier.
  4. Write in descending powers and record a, b, and c with their signs.

Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.

18. Pattern: get to standard form

Pattern

  1. Clear parentheses by distributing, and clear fractions by multiplying every term by the common denominator.
  2. Move all terms to one side so the other side is exactly zero.
  3. Choose the side that leaves the squared term positive - it makes factoring and the formula easier.
  4. Write in descending powers and record a, b, and c with their signs.

Do this before deciding which solving method to use. The coefficients you read here are what the decision depends on.

19. Check yourself: reading the coefficients

Check

Rewrite it in standard form on paper first, then choose.

\[ 5x^2 = 2x - 7 \]

Check your understanding

In standard form, what are a, b, and c?

  • A. a = 5, b = -2, c = 7 (correct)
  • B. a = 5, b = 2, c = -7
  • C. a = 5, b = -2, c = -7
  • D. a = 5, b = 2, c = 7

Answer: A

Why: Subtracting 2x and adding 7 to both sides gives 5x squared minus 2x plus 7 equals 0, so a is 5, b is negative 2, and c is positive 7. Both moved terms change sign because they crossed the equals sign.

Why B tempts people
Copied the terms straight across the equals sign without changing either sign - a subtraction on the right must become a subtraction on the left.
Why C tempts people
Changed the sign of the 2x term but left the 7 as a negative, forgetting that subtracting negative 7 on the right adds 7 on the left.
Why D tempts people
Changed the sign of the constant but not of the 2x term, so only half the move across the equals sign was done.

20. Factoring and the Zero-Product Property

Section

Part 2

21. The zero-product property

Concept

This one fact is the entire reason factoring solves equations.

\[ \text{If } A \cdot B = 0, \text{ then } A = 0 \text{ or } B = 0. \]

zero-product property — If a product of factors equals zero, at least one of those factors must itself be zero. It is true for zero and for no other number.

That last sentence is the whole trap. A product equal to twelve tells you nothing about the individual factors - many pairs of numbers multiply to twelve.

22. Take the definitions apart: standard form vs zero-product property

Definition probe

Sort into buckets

Every line below is part of the definition of standard form or of zero-product property — one or the other, never both. Put each where it belongs.

standard form
A quadratic written with every term on one side, in descending powers, and zero alone on the other side.; The three coefficients a, b, and c are read off directly from it - signs included.
zero-product property
If a product of factors equals zero, at least one of those factors must itself be zero.; It is true for zero and for no other number.
b1
A quadratic written with every term on one side, in descending powers, and zero alone on the other side. The three coefficients a, b, and c are read off directly from it - signs included.
b2
If a product of factors equals zero, at least one of those factors must itself be zero. It is true for zero and for no other number.

23. Factoring, then the zero-product property

Picture it

Animation

Shows: Factoring, then the zero-product property — a rendered Manim animation.

Rendered with Manim.

Takeaway: The right-hand side must be zero first, or the property does not apply.

24. Zero is the only number that talks

Intuition

Think of a receipt: price times quantity equals the total. If the total is zero dollars, you know something for certain - either the price was zero, or you bought zero items.

But if the total is twelve dollars, you know nothing specific: it could be one item at twelve dollars, two at six, or twelve at one.

Zero is the only product that forces a conclusion about the factors. That is why we insist on getting zero alone on one side.

25. Plan first: Worked example: solve by factoring

Step zero

Discussion prompt

Worked example: solve by factoring — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Find two numbers that multiply to 6 and add to 5

Answer:

  1. Find two numbers that multiply to 6 and add to 5
  2. Set each factor equal to zero
  3. Solve each linear equation
  4. Verify both solutions in the original equation

26. Worked example: solve by factoring

Worked example

\[ x^2 + 5x + 6 = 0 \]

It is already in standard form, so we may go straight to factoring.

Find two numbers that multiply to 6 and add to 5

Why: For a leading coefficient of one, the constant is the product of the two numbers in the binomials and the middle coefficient is their sum. Two and three work.

\[ (x + 2)(x + 3) = 0 \]

Set each factor equal to zero

Why: The product is zero, so by the zero-product property at least one factor is zero. Each possibility becomes its own small equation.

\[ x + 2 = 0 \quad \text{or} \quad x + 3 = 0 \]

Solve each linear equation

Why: One quadratic has become two one-step equations - that is the payoff of factoring.

\[ x = -2 \quad \text{or} \quad x = -3 \]

Verify both solutions in the original equation

Why: Substituting negative 2 gives 4 minus 10 plus 6, which is 0. Substituting negative 3 gives 9 minus 15 plus 6, which is also 0. Both check.

candidateleft side computedequals 0?
x = -24 - 10 + 6yes
x = -39 - 15 + 6yes

27. solve by factoring — line by line

Picture it

Animation

Shows: Each line of the worked example "solve by factoring", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Substituting negative 2 gives 4 minus 10 plus 6, which is 0. Substituting negative 3 gives 9 minus 15 plus 6, which is also 0. Both check.

28. Something is wrong here: setting factors equal to the constant

Anomaly

Predict first

A student writes this, and it looks reasonable:

The equation is not in standard form, but the left side factors, so it is tempting to split it right away.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: This feels like the same move - but the zero-product property says nothing about products that equal 8.

Move the constant across first so that zero is alone on one side.

Why: This feels like the same move - but the zero-product property says nothing about products that equal 8.

29. Trap: setting factors equal to the constant

Trap

The trap

The equation is not in standard form, but the left side factors, so it is tempting to split it right away.

\[ x^2 - 2x = 8 \]

Factor the left side and set each factor equal to 8

Why: This feels like the same move - but the zero-product property says nothing about products that equal 8.

\[ x(x - 2) = 8 \;\Rightarrow\; x = 8 \;\text{ or }\; x - 2 = 8 \]

Both candidates fail the original equation

Why: Neither number is a solution, and both true solutions were missed entirely.

candidateleft sideshould be 8
x = 864 - 16 = 48no
x = 10100 - 20 = 80no

The fix

Move the constant across first so that zero is alone on one side.

\[ x^2 - 2x = 8 \]

Subtract 8 from both sides, then factor

Why: Now the product really is zero, so the zero-product property applies.

\[ x^2 - 2x - 8 = 0 \;\Rightarrow\; (x - 4)(x + 2) = 0 \]

Solve each factor and check both

Why: Substituting 4 gives 16 minus 8, which is 8. Substituting negative 2 gives 4 plus 4, which is 8. Both check.

solutionleft sideshould be 8
x = 416 - 8 = 8yes
x = -24 + 4 = 8yes

30. Fill in: should be 8 for Trap: setting factors equal to the constant

Comparison

Comparison matrix

From Trap: setting factors equal to the constant: refill the should be 8 column from what you know. The rest of the table is as it appeared.

candidateleft sideshould be 8
x = 864 - 16 = 48no
x = 10100 - 20 = 80no

31. Guess the shape of the answer: Worked example: leading coefficient other…

Estimation

Predict first

With a leading coefficient of two, use the grouping method - sometimes called the ac method.

Commit before you compute: what does Worked example: leading coefficient other than one come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Verify both in the original equation

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. For one half: two times one quarter is one half, minus seven halves, plus three, which is zero.

32. Worked example: leading coefficient other than one

Worked example

\[ 2x^2 - 7x + 3 = 0 \]

With a leading coefficient of two, use the grouping method - sometimes called the ac method.

Multiply a times c, then find two numbers with that product and the middle sum

Why: Here a times c is 6 and the middle coefficient is negative 7. Negative 1 and negative 6 multiply to 6 and add to negative 7.

\[ ac = (2)(3) = 6, \qquad (-1)(-6) = 6, \quad -1 + (-6) = -7 \]

Split the middle term using those two numbers

Why: Negative 7x is rewritten as negative 6x minus x. The expression is unchanged - it is just written with four terms so it can be grouped.

\[ 2x^2 - 6x - x + 3 = 0 \]

Group in pairs and factor each pair

Why: The matching binomial left over in both pairs is what makes the grouping work; factor it out of the whole thing.

\[ 2x(x - 3) - 1(x - 3) = 0 \;\Rightarrow\; (2x - 1)(x - 3) = 0 \]

Apply the zero-product property

Why: Each factor gets its own equation, and each is linear.

\[ x = \tfrac{1}{2} \quad \text{or} \quad x = 3 \]

Verify both in the original equation

Why: For one half: two times one quarter is one half, minus seven halves, plus three, which is zero. For three: eighteen minus twenty-one plus three, which is zero.

candidatecomputationresult
x = 1/20.5 - 3.5 + 30
x = 318 - 21 + 30

33. leading coefficient other than one — line by line

Picture it

Animation

Shows: Each line of the worked example "leading coefficient other than one", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: For one half: two times one quarter is one half, minus seven halves, plus three, which is zero. For three: eighteen minus twenty-one plus three, which is zero.

34. Factor out the greatest common factor first

Concept

When every term shares a factor, pull it out before anything else. It makes the numbers smaller and sometimes finishes the factoring by itself.

\[ 3x^2 - 12x = 0 \;\Rightarrow\; 3x(x - 4) = 0 \]

Notice that the variable itself came out as a factor. That factor produces a genuine solution - and it is the one students throw away most often.

35. Complete the line: Worked example: when zero is a solution

Fill the middle

Fill in the blanks

From Worked example: when zero is a solution — finish the line. Write what belongs on the right of the equals sign before you look.

3x^2 = 12x

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Standard form first: zero must be alone on one side before any factor talk.

36. Worked example: when zero is a solution

Worked example

\[ 3x^2 = 12x \]

Both sides share a variable. Resist the urge to cancel it.

Subtract 12x from both sides

Why: Standard form first: zero must be alone on one side before any factor talk.

\[ 3x^2 - 12x = 0 \]

Factor out the greatest common factor

Why: Both terms contain a 3 and an x, so 3x comes out front.

\[ 3x(x - 4) = 0 \]

Set each factor equal to zero

Why: The constant 3 can never be zero, but the factor x can - and that gives the first solution.

\[ x = 0 \quad \text{or} \quad x = 4 \]

Verify both in the original equation

Why: With zero, both sides are 0. With four, the left is three times sixteen, or 48, and the right is twelve times four, also 48. Both check.

candidateleft sideright side
x = 000
x = 44848

37. when zero is a solution — line by line

Picture it

Animation

Shows: Each line of the worked example "when zero is a solution", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: With zero, both sides are 0. With four, the left is three times sixteen, or 48, and the right is twelve times four, also 48. Both check.

38. Something is wrong here: dividing both sides by the variable

Anomaly

Predict first

A student writes this, and it looks reasonable:

Cancelling the shared variable looks like a shortcut, and it does produce a true solution - just not all of them.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Dividing by a variable quietly assumes that variable is not zero - and here zero is exactly one of the answers.

Move everything to one side and factor instead of cancelling.

Why: Dividing by a variable quietly assumes that variable is not zero - and here zero is exactly one of the answers.

39. Trap: dividing both sides by the variable

Trap

The trap

Cancelling the shared variable looks like a shortcut, and it does produce a true solution - just not all of them.

\[ 3x^2 = 12x \]

Divide both sides by x

Why: Dividing by a variable quietly assumes that variable is not zero - and here zero is exactly one of the answers.

\[ 3x = 12 \;\Rightarrow\; x = 4 \]

One solution has vanished

Why: The value zero satisfies the original equation, since both sides become 0. Dividing by x erased it from the problem.

solutionfound?
x = 4yes
x = 0lost

The fix

Move everything to one side and factor instead of cancelling.

\[ 3x^2 = 12x \]

Subtract 12x and factor out 3x

Why: Factoring keeps the variable in the problem as a factor, where it can still produce a solution.

\[ 3x^2 - 12x = 0 \;\Rightarrow\; 3x(x - 4) = 0 \]

Both solutions survive and both check

Why: Zero makes the factor 3x zero, and four makes the factor x minus 4 zero. Substituting each into the original gives a true statement.

solutionfound?
x = 4yes
x = 0yes

40. What each one costs: Trap: dividing both sides by the variable

Trade off

Comparison matrix

From Trap: dividing both sides by the variable: every row here is a choice with a cost. Fill the found? column, then say which row you would actually pick and what you give up for it.

solutionfound?
x = 4yes
x = 0lost

41. What has to be given first: Worked example: a repeated solution

Missing information

Discussion prompt

Two numbers multiplying to 25 and adding to negative 10 - and they turn out to be the same number twice.

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Answer:

Negative 5 times negative 5 is 25, and negative 5 plus negative 5 is negative 10. The two identical factors collapse into a square.

42. Worked example: a repeated solution

Worked example

\[ x^2 - 10x + 25 = 0 \]

Two numbers multiplying to 25 and adding to negative 10 - and they turn out to be the same number twice.

Factor as a perfect square

Why: Negative 5 times negative 5 is 25, and negative 5 plus negative 5 is negative 10. The two identical factors collapse into a square.

\[ (x - 5)(x - 5) = 0 \;\Rightarrow\; (x - 5)^2 = 0 \]

Set the repeated factor equal to zero

Why: There is only one distinct equation to solve, so there is only one distinct solution.

\[ x = 5 \]

This is called a repeated or double solution. Graphically, the parabola touches the horizontal axis at exactly one point instead of crossing it twice.

Verify the solution in the original equation

Why: Twenty-five minus fifty plus twenty-five equals zero, so five checks. And it is the only value that does.

\[ 5^2 - 10(5) + 25 = 25 - 50 + 25 = 0 \]

43. a repeated solution — line by line

Picture it

Animation

Shows: Each line of the worked example "a repeated solution", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Twenty-five minus fifty plus twenty-five equals zero, so five checks. And it is the only value that does.

44. Check yourself: factoring after rearranging

Check

Get it to standard form first, then factor. Solve it on paper before choosing.

\[ x^2 = 3x + 10 \]

Check your understanding

What is the complete solution set?

  • A. x = 5 or x = -2 (correct)
  • B. x = -5 or x = 2
  • C. x = 10 or x = 13
  • D. x = 5 only

Answer: A

Why: Subtracting 3x and 10 gives x squared minus 3x minus 10 equals 0, which factors as the quantity x minus 5 times the quantity x plus 2. Checking: 25 equals 15 plus 10, and 4 equals negative 6 plus 10. Both work.

Why B tempts people
Reversed the signs inside the binomials, factoring as x plus 5 times x minus 2 - that product is x squared plus 3x minus 10, not the given equation.
Why C tempts people
Skipped standard form, factored the left side as x times x and set the pieces equal to 10 instead of zero, which the zero-product property does not allow.
Why D tempts people
Found the positive solution and stopped, assuming one answer finishes a quadratic; the negative factor gives a second, equally valid solution.

45. The Square Root Property

Section

Part 3

46. When there is no middle term

Concept

Some quadratics have no linear term at all. Those are the easiest equations in the whole unit - and factoring is not the fastest way to finish them.

\[ ax^2 + c = 0 \]

pure quadratic — A quadratic equation whose middle coefficient b is zero, so the only variable term is the squared one. Isolate the square and take a root instead of factoring.

47. Why a projectile is a parabola

Picture it

Animation

Shows: Why a projectile is a parabola — a rendered Manim animation.

Rendered with Manim.

Takeaway: Constant downward acceleration produces exactly a quadratic in time.

48. Picture it first: Two different numbers share one square

Picture it

Figure (svg): A number line with points at negative five and five, both marked as squaring to twenty-five

Undoing a square has to reach back to both of them.

Discussion prompt

Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.

Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.

Answer:

Squaring throws away the sign. A positive number and its negative twin land on exactly the same square.

49. Two different numbers share one square

Intuition

Squaring throws away the sign. A positive number and its negative twin land on exactly the same square.

Figure (svg): A number line with points at negative five and five, both marked as squaring to twenty-five

Undoing a square has to reach back to both of them.

So when you undo a square, you have to reach back for both originals. The plus-or-minus symbol is not decoration - it is the second solution.

50. The square root property

Concept

Once the squared quantity is alone on one side, you may take the root of both sides - as long as you keep both signs.

\[ \text{If } u^2 = k, \text{ then } u = \pm\sqrt{k}. \]

The letter standing in for the squared quantity can be a whole expression, not just a single variable. That is what makes this property so useful later.

\[ (x - 4)^2 = 18 \;\Rightarrow\; x - 4 = \pm\sqrt{18} \]

51. Where does each piece belong: Solving Quadratic Equations

Sorting

Sort into buckets

These are the pieces of Solving Quadratic Equations, out of order. Put each one back under the part of the lesson it belongs to.

What Makes an Equation Quadratic
Standard form is the starting line; The leading coefficient cannot be zero; Why a quadratic usually has two answers
Factoring and the Zero-Product Property
The zero-product property; Zero is the only number that talks; Worked example: solve by factoring
The Square Root Property
When there is no middle term; Two different numbers share one square; The square root property
s1
What Makes an Equation Quadratic is where Solving Quadratic Equations puts Standard form is the starting line, The leading coefficient cannot be zero, Why a quadratic usually has two answers. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.
s2
Factoring and the Zero-Product Property is where Solving Quadratic Equations puts The zero-product property, Zero is the only number that talks, Worked example: solve by factoring. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.
s3
The Square Root Property is where Solving Quadratic Equations puts When there is no middle term, Two different numbers share one square, The square root property. Knowing which part of the lesson a problem belongs to is most of knowing which method to reach for.

52. Worked example: isolate, then root

Worked example

\[ 3x^2 - 75 = 0 \]

There is no middle term, so go straight for the square root property.

Add 75 to both sides

Why: The squared term has to be alone before the square can be undone.

\[ 3x^2 = 75 \]

Divide both sides by 3

Why: The property applies to the square itself, not to three times the square. Strip the coefficient off first.

\[ x^2 = 25 \]

Take the square root of both sides, keeping both signs

Why: Two numbers square to twenty-five. Writing plus-or-minus captures both in one line.

\[ x = \pm 5 \]

Verify both solutions in the original equation

Why: Five squared and negative five squared are both twenty-five, so three times twenty-five minus seventy-five is zero either way.

candidateleft side computedequals 0?
x = 53(25) - 75 = 0yes
x = -53(25) - 75 = 0yes

53. isolate, then root — line by line

Picture it

Animation

Shows: Each line of the worked example "isolate, then root", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Five squared and negative five squared are both twenty-five, so three times twenty-five minus seventy-five is zero either way.

54. Trap: losing the negative solution

Trap

The trap

A calculator returns one root, so it is easy to record only one answer.

\[ x^2 = 49 \]

Write only the positive root

Why: The square root symbol by itself does mean the positive root - but the equation asks for every number whose square is forty-nine, not just that one.

\[ x = 7 \]

Half the answer is missing

Why: Negative seven squared is also forty-nine, so it satisfies the equation just as well and was never reported.

valuevalue squaredsolves the equation?
749yes
-749yes, but was dropped

The fix

Attach the plus-or-minus the instant you undo the square, before anything else happens.

\[ x^2 = 49 \]

Take the root of both sides with plus-or-minus

Why: The symbol is a promise to carry both branches through the rest of the problem.

\[ x = \pm 7 \]

Report both solutions and check both

Why: Both squares equal forty-nine, so both are genuine solutions of the original equation.

solutioncheckresult
x = 77 squared49
x = -7(-7) squared49

55. Break it on purpose: losing the negative solution

Break the constraint

Discussion prompt

The rule this trap just fixed:

The symbol is a promise to carry both branches through the rest of the problem.

Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?

Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.

Answer:

The square root symbol by itself does mean the positive root - but the equation asks for every number whose square is forty-nine, not just that one.

56. Plan first: Worked example: a squared binomial

Step zero

Discussion prompt

Worked example: a squared binomial — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Take the square root of both sides with plus-or-minus

Answer:

  1. Take the square root of both sides with plus-or-minus
  2. Simplify the radical
  3. Add 4 to both sides
  4. Verify both solutions in the original equation

57. Worked example: a squared binomial

Worked example

\[ (x - 4)^2 = 18 \]

The squared quantity is already alone, so the property applies immediately - to the whole binomial.

Take the square root of both sides with plus-or-minus

Why: The quantity being squared is the whole binomial, so that whole binomial is what equals the plus-or-minus root.

\[ x - 4 = \pm\sqrt{18} \]

Simplify the radical

Why: Eighteen contains a perfect square factor of nine, and the root of nine is three, which comes out front.

\[ \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \]

Add 4 to both sides

Why: Undo the subtraction last. The plus-or-minus stays attached to the radical, not to the four.

\[ x = 4 \pm 3\sqrt{2} \]

Verify both solutions in the original equation

Why: Subtracting four leaves plus or minus three root two, and squaring that gives nine times two, which is eighteen. Both branches check.

\[ \left(4 + 3\sqrt{2} - 4\right)^2 = \left(3\sqrt{2}\right)^2 = 9 \cdot 2 = 18 \]

58. a squared binomial — line by line

Picture it

Animation

Shows: Each line of the worked example "a squared binomial", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Subtracting four leaves plus or minus three root two, and squaring that gives nine times two, which is eighteen. Both branches check.

59. Worked example: when the square equals a negative

Worked example

\[ 2x^2 + 32 = 0 \]

Watch what happens when isolating the square leaves a negative number on the right.

Subtract 32, then divide by 2

Why: Same isolation moves as always - the sign of the result is what makes this one different.

\[ 2x^2 = -32 \;\Rightarrow\; x^2 = -16 \]

Note that no real number squares to a negative

Why: Squaring any real number, positive or negative, gives a result that is zero or positive. So there is no real solution here.

\[ x^2 = -16 \;\text{ has no real solution} \]

Take the root using the imaginary unit

Why: The imaginary unit is defined so that its square is negative one, which lets the root of a negative number be written down.

\[ x = \pm\sqrt{-16} = \pm 4i \]

Verify both solutions in the original equation

Why: Four times the imaginary unit, squared, is sixteen times negative one, which is negative sixteen. Doubling that gives negative thirty-two, and adding thirty-two gives zero.

\[ 2(4i)^2 + 32 = 2(16)(-1) + 32 = -32 + 32 = 0 \]

60. when the square equals a negative — line by line

Picture it

Animation

Shows: Each line of the worked example "when the square equals a negative", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Four times the imaginary unit, squared, is sixteen times negative one, which is negative sixteen. Doubling that gives negative thirty-two, and adding thirty-two gives zero.

61. Check yourself: the square root property

Check

Isolate the squared binomial first, then undo the square. Solve it on paper before choosing.

\[ 2(x - 5)^2 = 50 \]

Check your understanding

What is the complete solution set?

  • A. x = 10 or x = 0 (correct)
  • B. x = 10 only
  • C. x = 5 or x = -5
  • D. x = 10 or x = -10

Answer: A

Why: Dividing by 2 gives the squared binomial equal to 25, so the binomial equals plus or minus 5, and adding 5 gives 10 and 0. Checking: 2 times 25 is 50, and 2 times the square of negative 5 is also 50.

Why B tempts people
Took only the positive square root and dropped the plus-or-minus, which loses the branch that leads to zero.
Why C tempts people
Undid the square correctly but never added 5 back, reporting the value of the binomial instead of the value of x.
Why D tempts people
Applied the plus-or-minus to the finished answer rather than to the square root, negating 10 instead of negating the 5 before the final addition.

62. Completing the Square

Section

Part 4

63. Picture it first: The name is literal

Picture it

Figure (svg): A large square of side x, two rectangles of width three attached to its right and bottom, and a dashed nine by nine corner square that finishes the larger square

The missing corner is always the square of half the middle coefficient.

Discussion prompt

Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.

Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.

Answer:

Most quadratics do not factor and have no convenient missing middle term. Completing the square forces the square root property to apply by manufacturing a perfect square.

64. The name is literal

Intuition

Most quadratics do not factor and have no convenient missing middle term. Completing the square forces the square root property to apply by manufacturing a perfect square.

Figure (svg): A large square of side x, two rectangles of width three attached to its right and bottom, and a dashed nine by nine corner square that finishes the larger square

The missing corner is always the square of half the middle coefficient.

A square of side x with two strips of width three glued on is almost a bigger square. The only thing missing is the little corner - and its area is always the square of half the middle coefficient.

65. The perfect-square trinomial pattern

Concept

Every perfect-square trinomial has the same fingerprint: the constant is the square of half the middle coefficient.

\[ x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2 \]

expressionhalf the middle coefficientits squarefactors as
x squared plus 6x39(x + 3) squared
x squared minus 8x-416(x - 4) squared
x squared plus 5x5/225/4(x + 5/2) squared

Halve, then square. Notice the number inside the finished binomial is the halved value, not the squared one - that is where the sign and size come from.

66. Fill in: its square for The perfect-square trinomial pattern

Comparison

Comparison matrix

From The perfect-square trinomial pattern: refill the its square column from what you know. The rest of the table is as it appeared.

expressionhalf the middle coefficientits squarefactors as
x squared plus 6x39(x + 3) squared
x squared minus 8x-416(x - 4) squared
x squared plus 5x5/225/4(x + 5/2) squared

67. State the rule before it runs: Worked example: completing the square

Hypothesis

Predict first

Worked example: completing the square is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.

Correct: Move the constant to the right side

Why: Completing the square operates on the two variable terms; the constant has to get out of the way first.

A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.

68. Worked example: completing the square

Worked example

\[ x^2 + 6x + 5 = 0 \]

This one does factor, which makes it a safe place to watch the method work.

Move the constant to the right side

Why: Completing the square operates on the two variable terms; the constant has to get out of the way first.

\[ x^2 + 6x = -5 \]

Halve the middle coefficient and square it

Why: Half of six is three, and three squared is nine. That nine is the corner piece the square is missing.

\[ \left(\frac{6}{2}\right)^2 = 3^2 = 9 \]

Add 9 to both sides

Why: Adding to one side alone would change the equation. Adding to both keeps it equivalent while building the square.

\[ x^2 + 6x + 9 = -5 + 9 = 4 \]

Factor the left side and apply the square root property

Why: The left side is now a perfect square, so the whole binomial equals plus or minus the root of four.

\[ (x + 3)^2 = 4 \;\Rightarrow\; x + 3 = \pm 2 \]

Subtract 3 from both sides on each branch

Why: Negative three plus two is negative one; negative three minus two is negative five.

\[ x = -1 \quad \text{or} \quad x = -5 \]

Verify both solutions in the original equation

Why: One minus six plus five is zero, and twenty-five minus thirty plus five is zero. Both check.

candidateleft side computedequals 0?
x = -11 - 6 + 5yes
x = -525 - 30 + 5yes

69. Completing the square

Picture it

Animation

Shows: Completing the square — a rendered Manim animation.

Rendered with Manim.

Takeaway: Add the square of half the middle coefficient, to both sides.

70. Worked example: an answer no factoring would find

Worked example

\[ x^2 - 8x + 3 = 0 \]

No pair of integers multiplies to three and adds to negative eight, so factoring is off the table. Completing the square does not care.

Subtract 3 from both sides

Why: Clear the constant off the variable side so the square can be built there.

\[ x^2 - 8x = -3 \]

Halve negative 8 and square the result

Why: Half of negative eight is negative four, and negative four squared is positive sixteen. The added constant is always positive.

\[ \left(\frac{-8}{2}\right)^2 = (-4)^2 = 16 \]

Add 16 to both sides and factor

Why: The left side becomes the square of x minus four, using the halved value with its sign. On the right, negative three plus sixteen is thirteen.

\[ (x - 4)^2 = 13 \]

Apply the square root property and isolate x

Why: Thirteen has no perfect-square factor, so the radical stays as it is and the answer is exact.

\[ x - 4 = \pm\sqrt{13} \;\Rightarrow\; x = 4 \pm \sqrt{13} \]

Verify by expanding the solution back into the original equation

Why: Expanding the square gives sixteen plus eight root thirteen plus thirteen; subtracting eight times the solution removes thirty-two and eight root thirteen; adding three leaves zero exactly.

\[ (4 + \sqrt{13})^2 - 8(4 + \sqrt{13}) + 3 = 29 + 8\sqrt{13} - 32 - 8\sqrt{13} + 3 = 0 \]

71. an answer no factoring would find — line by line

Picture it

Animation

Shows: Each line of the worked example "an answer no factoring would find", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Expanding the square gives sixteen plus eight root thirteen plus thirteen; subtracting eight times the solution removes thirty-two and eight root thirteen; adding three leaves zero exactly.

72. A leading coefficient other than one

Concept

The halve-and-square rule assumes the squared term has a coefficient of one. If it does not, make it one first.

\[ ax^2 + bx + c = 0 \;\Rightarrow\; x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]

Divide every single term by the leading coefficient - all three of them, on both sides. Dividing one term is the single most common way this method goes wrong.

73. What has to happen first: Worked example: divide through first

Ranking

Put in order

Put the moves of Worked example: divide through first into the order they have to happen.

  1. Divide every term by 2
  2. Move the constant across
  3. Halve 4, square it, and add to both sides
  4. Take the root and solve both branches
  5. Verify both solutions in the original equation

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Each of the three terms is halved.

74. Worked example: divide through first

Worked example

\[ 2x^2 + 8x - 10 = 0 \]

The leading coefficient is two, so the very first move is to divide the whole equation by two.

Divide every term by 2

Why: Each of the three terms is halved. The right side is zero, and zero divided by two is still zero.

\[ x^2 + 4x - 5 = 0 \]

Move the constant across

Why: Add five to both sides so only the variable terms remain on the left.

\[ x^2 + 4x = 5 \]

Halve 4, square it, and add to both sides

Why: Half of four is two, and two squared is four. Five plus four is nine on the right.

\[ x^2 + 4x + 4 = 9 \;\Rightarrow\; (x + 2)^2 = 9 \]

Take the root and solve both branches

Why: Negative two plus three is one; negative two minus three is negative five.

\[ x + 2 = \pm 3 \;\Rightarrow\; x = 1 \quad \text{or} \quad x = -5 \]

Verify both solutions in the original equation

Why: For one: two plus eight minus ten is zero. For negative five: fifty minus forty minus ten is zero. Both check in the equation we started with, not the divided one.

candidatecomputationresult
x = 12(1) + 8(1) - 100
x = -52(25) + 8(-5) - 100

75. divide through first — line by line

Picture it

Animation

Shows: Each line of the worked example "divide through first", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: For one: two plus eight minus ten is zero. For negative five: fifty minus forty minus ten is zero. Both check in the equation we started with, not the divided one.

76. Something is wrong here: dividing only the squared term

Anomaly

Predict first

A student writes this, and it looks reasonable:

The goal is a leading coefficient of one, so it is tempting to just change the leading coefficient and move on.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: This looks like it achieved the goal, but the equation is no longer equivalent - two of the terms were never touched.

Divide the entire equation - every term on both sides - by the leading coefficient.

Why: This looks like it achieved the goal, but the equation is no longer equivalent - two of the terms were never touched.

77. Trap: dividing only the squared term

Trap

The trap

The goal is a leading coefficient of one, so it is tempting to just change the leading coefficient and move on.

\[ 3x^2 + 12x - 15 = 0 \]

Divide only the squared term by 3

Why: This looks like it achieved the goal, but the equation is no longer equivalent - two of the terms were never touched.

\[ x^2 + 12x - 15 = 0 \;\Rightarrow\; (x + 6)^2 = 51 \]

The answers do not satisfy the original equation

Why: The square root of fifty-one is about 7.14, so the candidates are about 1.14 and about negative 13.14. Substituting the first into the original gives about 2.6, not zero.

candidateoriginal left sideequals 0?
about 1.14about 2.6no
about -13.14about 345.4no

The fix

Divide the entire equation - every term on both sides - by the leading coefficient.

\[ 3x^2 + 12x - 15 = 0 \]

Divide all three terms by 3

Why: Dividing an entire equation by a nonzero number leaves an equivalent equation with the same solutions.

\[ x^2 + 4x - 5 = 0 \;\Rightarrow\; (x + 2)^2 = 9 \]

Solve and verify in the original

Why: The solutions are one and negative five. Three plus twelve minus fifteen is zero, and seventy-five minus sixty minus fifteen is zero. Both check.

solutionoriginal left sideequals 0?
x = 13 + 12 - 15yes
x = -575 - 60 - 15yes

78. Decode the notation: Trap: dividing only the squared term

Notation

Annotate

From Trap: dividing only the squared term — read this one piece at a time. What is each part doing?

On: \( 3x^2 + 12x - 15 = 0 \)

  • This looks like it achieved the goal, but the equation is no longer equivalent - two of the terms were never touched.
  • The square root of fifty-one is about 7.14, so the candidates are about 1.14 and about negative 13.14. Substituting the first into the original gives about 2.6, not zero.
  • Dividing an entire equation by a nonzero number leaves an equivalent equation with the same solutions.

79. Guess the shape of the answer: Worked example: fractions along the way

Estimation

Predict first

Dividing by three makes a fraction appear. That is normal - keep going.

Commit before you compute: what does Worked example: fractions along the way come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Verify by rebuilding the original expression

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Three times the square of x minus two, minus ten, is exactly the original left side.

80. Worked example: fractions along the way

Worked example

\[ 3x^2 - 12x + 2 = 0 \]

Dividing by three makes a fraction appear. That is normal - keep going.

Divide every term by 3

Why: Twelve divided by three is four exactly; two divided by three stays as a fraction, which is fine.

\[ x^2 - 4x + \frac{2}{3} = 0 \]

Move the fraction to the right side

Why: Only the two variable terms belong on the left while the square is built.

\[ x^2 - 4x = -\frac{2}{3} \]

Halve negative 4, square it, add to both sides

Why: Half of negative four is negative two, whose square is four. On the right, four minus two thirds is ten thirds.

\[ (x - 2)^2 = 4 - \frac{2}{3} = \frac{10}{3} \]

Take the root and rationalize the denominator

Why: The root of ten thirds becomes the root of thirty over three once the denominator is rationalized.

\[ x - 2 = \pm\sqrt{\frac{10}{3}} = \pm\frac{\sqrt{30}}{3} \;\Rightarrow\; x = 2 \pm \frac{\sqrt{30}}{3} \]

Verify by rebuilding the original expression

Why: Three times the square of x minus two, minus ten, is exactly the original left side. Since that square is ten thirds, three times it is ten, and ten minus ten is zero.

\[ 3(x-2)^2 - 10 = 3\left(\frac{10}{3}\right) - 10 = 0, \qquad 3(x-2)^2 - 10 = 3x^2 - 12x + 2 \]

81. fractions along the way — line by line

Picture it

Animation

Shows: Each line of the worked example "fractions along the way", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Three times the square of x minus two, minus ten, is exactly the original left side. Since that square is ten thirds, three times it is ten, and ten minus ten is zero.

82. Pattern: complete the square

Pattern

  1. Write the equation in standard form, then divide every term by the leading coefficient so the squared term has coefficient one.
  2. Move the constant to the other side, leaving only the squared and linear terms together.
  3. Halve the middle coefficient, square that half, and add it to both sides.
  4. Factor the left side as a squared binomial - the number inside is the halved value, sign included.
  5. Apply the square root property with plus-or-minus, then isolate the variable.

Step three is the only new idea here. Everything else is moves you already own.

This method always works, even when factoring fails - and running it on the general equation is exactly how the quadratic formula is born.

83. Where does it stop working: Pattern: complete the square

Edge cases

Discussion prompt

Pattern: complete the square works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".

Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.

Answer:

Step three is the only new idea here. Everything else is moves you already own.

84. Check yourself: what completes the square

Check

Find the constant that turns this into a perfect-square trinomial.

\[ x^2 - 10x + \underline{\phantom{00}} \]

Check your understanding

Which number completes the square?

  • A. 25 (correct)
  • B. 100
  • C. 5
  • D. -25

Answer: A

Why: Half of negative 10 is negative 5, and negative 5 squared is positive 25. The result factors as the square of the quantity x minus 5, which expands back to x squared minus 10x plus 25.

Why B tempts people
Squared the middle coefficient without halving it first, giving 100 instead of 25.
Why C tempts people
Halved the middle coefficient but stopped there and never squared the result.
Why D tempts people
Kept the negative sign after squaring; squaring negative 5 gives positive 25, and a perfect-square trinomial's constant is never negative.

85. The Quadratic Formula

Section

Part 5

86. Do the work once, keep it forever

Concept

Completing the square works on every quadratic - which means the steps never change. Only the numbers do.

So run the method one final time on the general equation, with letters instead of numbers. The result is a formula that answers every quadratic at once.

That is all the quadratic formula is: completing the square, done in advance, so you never have to do it again.

87. Complete the line: Worked example: deriving the formula

Fill the middle

Fill in the blanks

From Worked example: deriving the formula — finish the line. Write what belongs on the right of the equals sign before you look.

x^2 + \frac-\frac{c}{a}___x = ___

Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The leading coefficient must be one before halving and squaring.

88. Worked example: deriving the formula

Worked example

\[ ax^2 + bx + c = 0, \qquad a \ne 0 \]

Same five steps as before. Watch the letters go through them.

Divide every term by a

Why: The leading coefficient must be one before halving and squaring. Every term is divided, not just the first.

\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]

Move the constant term to the right side

Why: Clear the variable side so the square can be assembled there.

\[ x^2 + \frac{b}{a}x = -\frac{c}{a} \]

Halve the middle coefficient, square it, add to both sides

Why: Half of b over a is b over 2a, and its square is b squared over 4a squared. Add that to both sides.

\[ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a} \]

Factor the left and combine the right over a common denominator

Why: The left is a perfect square with the halved value inside. On the right, c over a becomes 4ac over 4a squared.

\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \]

Apply the square root property with plus-or-minus

Why: The root of 4a squared in the denominator is 2a, and the numerator keeps its radical because it will not simplify in general.

\[ x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a} \]

Subtract the fraction and combine over the shared denominator

Why: Both pieces already have denominator 2a, so they merge into a single fraction.

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Check the formula against an equation we already solved

Why: For x squared plus 5x plus 6, the coefficients are 1, 5, and 6, so the radical holds 25 minus 24, which is 1. That gives negative 5 plus or minus 1, over 2 - namely negative 2 and negative 3, exactly the factoring answers from Part 2.

\[ x = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2} = -2 \;\text{ or }\; -3 \]

89. deriving the formula — line by line

Picture it

Animation

Shows: Each line of the worked example "deriving the formula", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: For x squared plus 5x plus 6, the coefficients are 1, 5, and 6, so the radical holds 25 minus 24, which is 1. That gives negative 5 plus or minus 1, over 2 - namely negative 2 and negative 3, exactly the factoring answers from Part 2.

90. The formula, and what to memorize

Concept

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Three details students lose points on, every term:

quadratic formula — The solution of the general quadratic in standard form, obtained by completing the square on it once. It works on every quadratic, factorable or not, real solutions or complex.

91. Plan first: Worked example: the formula on a friendly quadratic

Step zero

Discussion prompt

Worked example: the formula on a friendly quadratic — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: List a, b, and c with their signs

Answer:

  1. List a, b, and c with their signs
  2. Substitute into the formula, keeping every value in parentheses
  3. Simplify under the radical
  4. Split into the two branches
  5. Verify both solutions in the original equation

92. Worked example: the formula on a friendly quadratic

Worked example

\[ 2x^2 + 5x - 3 = 0 \]

Already in standard form, so read the three coefficients straight off it.

List a, b, and c with their signs

Why: Writing them down separately before substituting is what stops sign errors. The constant is negative three, not three.

\[ a = 2, \quad b = 5, \quad c = -3 \]

Substitute into the formula, keeping every value in parentheses

Why: Parentheses protect the signs while the arithmetic happens - especially the negative c inside the product.

\[ x = \frac{-(5) \pm \sqrt{(5)^2 - 4(2)(-3)}}{2(2)} \]

Simplify under the radical

Why: Twenty-five minus a negative twenty-four becomes twenty-five plus twenty-four, which is forty-nine - a perfect square.

\[ x = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4} \]

Split into the two branches

Why: Negative five plus seven is two, over four, which reduces to one half. Negative five minus seven is negative twelve, over four, which is negative three.

\[ x = \frac{1}{2} \quad \text{or} \quad x = -3 \]

Verify both solutions in the original equation

Why: For one half: two times one quarter is one half, plus five halves, minus three, which is zero. For negative three: eighteen minus fifteen minus three, which is zero.

candidatecomputationresult
x = 1/20.5 + 2.5 - 30
x = -318 - 15 - 30

93. The formula IS completing the square

Picture it

Animation

Shows: The formula IS completing the square — a rendered Manim animation.

Rendered with Manim.

Takeaway: Do the general case once and you never do it again.

94. Substituting negatives safely

Concept

Almost every quadratic-formula error is a sign error, and almost every sign error comes from writing a negative coefficient without parentheses.

\[ b = -5 \;\Rightarrow\; -b = -(-5) = 5, \qquad b^2 = (-5)^2 = 25 \]

Two habits fix this permanently: write the coefficients on their own line first, and put every substituted value inside parentheses before simplifying anything.

95. Worked example: negative b and negative c

Worked example

\[ 3x^2 - 5x - 2 = 0 \]

Two of the three coefficients are negative. Write them down before touching the formula.

List the coefficients with their signs

Why: The minus signs belong to the numbers themselves. Recording them here is what makes the substitution safe.

\[ a = 3, \quad b = -5, \quad c = -2 \]

Substitute with parentheses everywhere

Why: Negative b becomes the opposite of negative five, which is positive five. And b squared is the square of negative five, which is positive twenty-five.

\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} \]

Simplify the discriminant

Why: Four times three times negative two is negative twenty-four; subtracting negative twenty-four adds it. Twenty-five plus twenty-four is forty-nine.

\[ x = \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6} \]

Split into the two branches and reduce

Why: Five plus seven is twelve, over six, which is two. Five minus seven is negative two, over six, which reduces to negative one third.

\[ x = 2 \quad \text{or} \quad x = -\frac{1}{3} \]

Verify both solutions in the original equation

Why: For two: twelve minus ten minus two is zero. For negative one third: three times one ninth is one third, minus five times negative one third adds five thirds, and one third plus five thirds minus two is zero.

candidatecomputationresult
x = 23(4) - 5(2) - 20
x = -1/31/3 + 5/3 - 20

96. negative b and negative c — line by line

Picture it

Animation

Shows: Each line of the worked example "negative b and negative c", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: For two: twelve minus ten minus two is zero. For negative one third: three times one ninth is one third, minus five times negative one third adds five thirds, and one third plus five thirds minus two is zero.

97. Something is wrong here: dropping a coefficient's negative sign

Anomaly

Predict first

A student writes this, and it looks reasonable:

The equation shows a minus in front of the middle term, and it is easy to copy the number without it.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: The minus was read as an operation between terms rather than as part of the coefficient, so the leading minus in the formula is applied to a positive six.

The sign in front of a term belongs to that term's coefficient. Record it before substituting.

Why: The minus was read as an operation between terms rather than as part of the coefficient, so the leading minus in the formula is applied to a positive six.

98. Trap: dropping a coefficient's negative sign

Trap

The trap

The equation shows a minus in front of the middle term, and it is easy to copy the number without it.

\[ x^2 - 6x + 7 = 0 \]

Substitute b as positive 6

Why: The minus was read as an operation between terms rather than as part of the coefficient, so the leading minus in the formula is applied to a positive six.

\[ x = \frac{-6 \pm \sqrt{36 - 28}}{2} = \frac{-6 \pm 2\sqrt{2}}{2} = -3 \pm \sqrt{2} \]

Neither candidate satisfies the original equation

Why: Substituting negative three plus root two gives thirty-six minus twelve root two, which is about 19.0 - nowhere near zero. Both answers are the exact negatives of the true ones.

candidateoriginal left sideequals 0?
about -1.586about 19.0no
about -4.414about 52.97no

The fix

The sign in front of a term belongs to that term's coefficient. Record it before substituting.

\[ x^2 - 6x + 7 = 0, \qquad a = 1, \; b = -6, \; c = 7 \]

Substitute b as negative 6 inside parentheses

Why: The opposite of negative six is positive six, and the square of negative six is positive thirty-six. Only the numerator's leading sign changes.

\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(7)}}{2(1)} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2} \]

Both solutions check exactly

Why: Expanding gives nine plus six root two plus two, minus eighteen minus six root two, plus seven - the radical terms cancel and the numbers total zero.

\[ (3 + \sqrt{2})^2 - 6(3 + \sqrt{2}) + 7 = 11 + 6\sqrt{2} - 18 - 6\sqrt{2} + 7 = 0 \]

99. Say it in words: Trap: dropping a coefficient's negative sign

Translation

\( x^2 - 6x + 7 = 0, \qquad a = 1, \; b = -6, \; c = 7 \)

Draw it

Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.

100. What has to be given first: Worked example: an irrational answer, fully…

Missing information

Discussion prompt

This one does not factor, and the discriminant is not a perfect square. The formula handles it anyway.

What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.

Hint: Anything you would have to invent to get started is a thing the problem must supply.

Answer:

a is 2, b is negative 4, c is negative 3. Parentheses keep both minus signs intact.

101. Worked example: an irrational answer, fully reduced

Worked example

\[ 2x^2 - 4x - 3 = 0 \]

This one does not factor, and the discriminant is not a perfect square. The formula handles it anyway.

List the coefficients and substitute

Why: a is 2, b is negative 4, c is negative 3. Parentheses keep both minus signs intact.

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-3)}}{2(2)} \]

Simplify the discriminant

Why: Sixteen minus negative twenty-four is sixteen plus twenty-four, which is forty.

\[ x = \frac{4 \pm \sqrt{40}}{4} \]

Simplify the radical

Why: Forty is four times ten, and the root of four is two, so a two comes out of the radical.

\[ \sqrt{40} = 2\sqrt{10} \;\Rightarrow\; x = \frac{4 \pm 2\sqrt{10}}{4} \]

Factor 2 out of the numerator and reduce

Why: Every term in the numerator is divisible by two, so the common factor comes out and cancels one factor of two in the denominator.

\[ x = \frac{2\left(2 \pm \sqrt{10}\right)}{4} = \frac{2 \pm \sqrt{10}}{2} \]

Verify the solution in the original equation

Why: Twice the square of the answer is seven plus two root ten, and four times the answer is four plus two root ten. Subtracting and then subtracting three leaves exactly zero.

\[ 2x^2 - 4x - 3 = \left(7 + 2\sqrt{10}\right) - \left(4 + 2\sqrt{10}\right) - 3 = 0 \]

102. an irrational answer, fully reduced — line by line

Picture it

Animation

Shows: Each line of the worked example "an irrational answer, fully reduced", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Twice the square of the answer is seven plus two root ten, and four times the answer is four plus two root ten. Subtracting and then subtracting three leaves exactly zero.

103. Something is wrong here: cancelling into only one term

Anomaly

Predict first

A student writes this, and it looks reasonable:

A number appears in the numerator and again in the denominator, so it looks cancellable on sight.

It is wrong. Say what breaks — and say it before you turn the page.

Correct: Cancelling requires a factor of the entire numerator.

Factor the common piece out of the whole numerator first, then cancel the factor.

Why: Cancelling requires a factor of the entire numerator. The 4 here is only one of two added terms, so it is not a factor of the whole thing.

104. Trap: cancelling into only one term

Trap

The trap

A number appears in the numerator and again in the denominator, so it looks cancellable on sight.

\[ x = \frac{4 \pm 2\sqrt{10}}{4} \]

Strike the 4 in the numerator against the 4 below

Why: Cancelling requires a factor of the entire numerator. The 4 here is only one of two added terms, so it is not a factor of the whole thing.

\[ x = 1 \pm 2\sqrt{10} \]

The result is off by a wide margin

Why: Substituting one plus two root ten into the original left side gives exactly seventy-five, not zero. The true root is about 2.581; this claims about 7.325.

candidateoriginal left sideequals 0?
about 7.32575no
about -5.32575no

The fix

Factor the common piece out of the whole numerator first, then cancel the factor.

\[ x = \frac{4 \pm 2\sqrt{10}}{4} \]

Factor 2 out of both numerator terms

Why: Now the two really is a factor of the entire numerator, which is what makes cancelling legal.

\[ x = \frac{2\left(2 \pm \sqrt{10}\right)}{4} = \frac{2 \pm \sqrt{10}}{2} \]

The reduced answer checks in the original equation

Why: Twice the square of this value is seven plus two root ten and four times it is four plus two root ten, so the expression collapses to zero exactly.

candidateoriginal left sideequals 0?
about 2.5810yes
about -0.5810yes

105. Which of these survive contact with Solving Quadratic Equations?

Two truths and a lie

Sort into buckets

Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.

Holds up
An equation is quadratic when you can rearrange it so one side is zero and the other side has a squared variable as its highest power.; The squared term is what makes it quadratic. If its coefficient is zero, the squared term disappears and you are back to a linear equation.; Graph the left side and you get a parabola - a U-shaped curve. Solving the equation means asking where that curve sits at height zero.
Breaks
The equation is not in standard form, but the left side factors, so it is tempting to split it right away.; Cancelling the shared variable looks like a shortcut, and it does produce a true solution - just not all of them.
sound
These are stated as this lesson states them — each one survives the edge cases Solving Quadratic Equations puts it through.
flawed
Each of these is lifted from a trap in this deck: reasonable-sounding, and wrong in a way that only shows up once you rely on it.

106. Check yourself: the formula with negative coefficients

Check

Write down the three coefficients with their signs, then substitute with parentheses.

\[ 2x^2 - 3x - 2 = 0 \]

Check your understanding

What are the solutions?

  • A. x = 2 or x = -1/2 (correct)
  • B. x = -2 or x = 1/2
  • C. No real solutions
  • D. x = 4 or x = -1

Answer: A

Why: With a equal to 2, b equal to negative 3, and c equal to negative 2, the discriminant is 9 plus 16, which is 25. The formula gives 3 plus or minus 5, all over 4, so x is 2 or negative one half. Checking 2: 8 minus 6 minus 2 is 0.

Why B tempts people
Used negative 3 in the numerator instead of the opposite of b; the formula begins with the opposite of b, so a negative b becomes positive.
Why C tempts people
Computed the discriminant as 9 minus 16 by ignoring the negative sign on c, producing a negative value and a false conclusion that nothing works.
Why D tempts people
Divided by a instead of by two times a, using a denominator of 2 rather than 4.

107. The Discriminant

Section

Part 6

108. The piece under the radical has a name

Concept

Everything interesting in the quadratic formula happens under the radical. That expression gets its own name.

\[ D = b^2 - 4ac \]

discriminant — The quantity b squared minus 4ac from inside the quadratic formula's radical. Its sign alone tells you how many solutions there are and whether they are real, before you finish solving.

109. Teach it back: The piece under the radical has a name

Explain it

Discussion prompt

Explain The piece under the radical has a name to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.

Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.

Answer:

Everything interesting in the quadratic formula happens under the radical. That expression gets its own name.

110. One number, computed first, saves the whole problem

Intuition

Think of the discriminant as a weather report you read before leaving the house. It does not do the trip for you, but it tells you what to pack.

If it comes out positive, expect two different answers. If it comes out zero, expect one. If it comes out negative, expect the imaginary unit to show up - and stop looking for real answers.

Computing it first also catches arithmetic slips early, while there is still only one small number to recheck.

111. By analogy: One number, computed first, saves the whole problem

Analogy

Discussion prompt

Explain One number, computed first, saves the whole problem by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.

Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.

Answer:

Think of the discriminant as a weather report you read before leaving the house. It does not do the trip for you, but it tells you what to pack.

112. Picture it first: The three cases

Picture it

Figure (svg): Three parabolas above a horizontal axis: the first crossing it twice, the second touching it once, the third staying entirely above it

The discriminant is the algebra behind these three pictures.

Discussion prompt

Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.

Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.

Answer:

One more free fact: when the discriminant is a perfect square, the radical disappears, the answers are rational, and the equation would have factored.

113. The three cases

Concept

discriminantnumber and type of solutionswhat the parabola does
positivetwo different real solutionscrosses the horizontal axis twice
zeroone repeated real solutiontouches the axis at exactly one point
negativetwo complex conjugate solutionsnever reaches the axis

Figure (svg): Three parabolas above a horizontal axis: the first crossing it twice, the second touching it once, the third staying entirely above it

The discriminant is the algebra behind these three pictures.

One more free fact: when the discriminant is a perfect square, the radical disappears, the answers are rational, and the equation would have factored.

114. What each one costs: The three cases

Trade off

Comparison matrix

From The three cases: every row here is a choice with a cost. Fill the what the parabola does column, then say which row you would actually pick and what you give up for it.

discriminantnumber and type of solutionswhat the parabola does
positivetwo different real solutionscrosses the horizontal axis twice
zeroone repeated real solutiontouches the axis at exactly one point
negativetwo complex conjugate solutionsnever reaches the axis

115. The discriminant, seen

Picture it

Animation

Shows: The discriminant, seen — a rendered Manim animation.

Rendered with Manim.

Takeaway: Two crossings, one touch, or none — decided entirely by b squared minus 4ac.

116. Worked example: a discriminant of zero

Worked example

\[ 4x^2 - 12x + 9 = 0 \]

Predict the answer count before solving.

List the coefficients and compute the discriminant

Why: a is 4, b is negative 12, c is 9. The square of negative twelve is 144, and four times four times nine is also 144.

\[ D = (-12)^2 - 4(4)(9) = 144 - 144 = 0 \]

Predict one repeated real solution

Why: A discriminant of zero makes the plus-or-minus branch add and subtract nothing, so the two branches land on the same number.

\[ x = \frac{-b \pm 0}{2a} = \frac{-b}{2a} \]

Finish the formula

Why: The opposite of negative twelve is twelve, and two times four is eight. Twelve over eight reduces to three halves.

\[ x = \frac{12}{8} = \frac{3}{2} \]

Verify the solution in the original equation

Why: Three halves squared is nine fourths; four times that is nine. Twelve times three halves is eighteen. Nine minus eighteen plus nine is zero, and the expression is the perfect square of the quantity two x minus three.

\[ 4\left(\tfrac{9}{4}\right) - 12\left(\tfrac{3}{2}\right) + 9 = 9 - 18 + 9 = 0 \]

117. a discriminant of zero — line by line

Picture it

Animation

Shows: Each line of the worked example "a discriminant of zero", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Three halves squared is nine fourths; four times that is nine. Twelve times three halves is eighteen. Nine minus eighteen plus nine is zero, and the expression is the perfect square of the quantity two x minus three.

118. Guess the shape of the answer: Worked example: a negative discriminant

Estimation

Predict first

Nothing multiplies to five and adds to two, so factoring is out. Check the discriminant before committing.

Commit before you compute: what does Worked example: a negative discriminant come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.

Correct: Verify one solution in the original equation

Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Squaring negative one plus two i gives one minus four i plus four times negative one, which is negative three minus four i.

119. Worked example: a negative discriminant

Worked example

\[ x^2 + 2x + 5 = 0 \]

Nothing multiplies to five and adds to two, so factoring is out. Check the discriminant before committing.

Compute the discriminant

Why: a is 1, b is 2, c is 5. Four minus twenty is negative sixteen, so there are no real solutions.

\[ D = 2^2 - 4(1)(5) = 4 - 20 = -16 \]

Substitute into the formula and write the negative root with the imaginary unit

Why: The root of negative sixteen is four times the imaginary unit, since the imaginary unit squares to negative one.

\[ x = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} \]

Divide every term of the numerator by 2

Why: Both the real part and the imaginary part are divided - the same all-terms rule as everywhere else.

\[ x = -1 \pm 2i \]

Verify one solution in the original equation

Why: Squaring negative one plus two i gives one minus four i plus four times negative one, which is negative three minus four i. Adding twice the solution gives negative two plus four i, and adding five clears everything to zero. Its conjugate checks the same way.

\[ (-1+2i)^2 + 2(-1+2i) + 5 = (-3-4i) + (-2+4i) + 5 = 0 \]

120. a negative discriminant — line by line

Picture it

Animation

Shows: Each line of the worked example "a negative discriminant", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Squaring negative one plus two i gives one minus four i plus four times negative one, which is negative three minus four i. Adding twice the solution gives negative two plus four i, and adding five clears everything to zero. Its conjugate checks the same way.

121. Rule out three: Check yourself: reading the discriminant

Elimination

Eliminate the wrong options

How many solutions does this equation have, and of what type?

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

  • A. Two different real solutions
  • B. One repeated real solution
  • C. Two complex conjugate solutions
  • D. No solutions of any kind

Survives elimination: C

Why: The discriminant is the square of negative 4 minus 4 times 3 times 2, which is 16 minus 24, or negative 8. A negative discriminant makes the radical imaginary, so the two solutions are complex conjugates and the parabola never touches the horizontal axis.

122. Check yourself: reading the discriminant

Check

Compute the discriminant only - do not solve the equation.

\[ 3x^2 - 4x + 2 = 0 \]

Check your understanding

How many solutions does this equation have, and of what type?

  • A. Two different real solutions
  • B. One repeated real solution
  • C. Two complex conjugate solutions (correct)
  • D. No solutions of any kind

Answer: C

Why: The discriminant is the square of negative 4 minus 4 times 3 times 2, which is 16 minus 24, or negative 8. A negative discriminant makes the radical imaginary, so the two solutions are complex conjugates and the parabola never touches the horizontal axis.

Why A tempts people
Added 4ac instead of subtracting it, getting 40 and a false positive discriminant; c is positive here, so the 24 must be subtracted.
Why B tempts people
Assumed that any discriminant which is not a perfect square collapses the two roots into one; the repeated case happens only when the discriminant is exactly zero.
Why D tempts people
Stopped at the negative discriminant and concluded nothing works. Over the complex numbers the square root of a negative number does exist, giving two conjugate solutions.

123. Choosing a Method and Applying It

Section

Part 7

124. Without one step: Pattern: pick the fastest method

Constraint

Discussion prompt

Run Pattern: pick the fastest method with this step confiscated:

If neither shortcut is there, compute the discriminant and use the quadratic formula.

Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.

Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.

Answer:

  1. Put the equation in standard form and read a, b, and c with their signs.
  2. Scan for a missing middle term or an obvious factorization - those are the shortcuts.
  3. If neither shortcut is there, compute the discriminant and use the quadratic formula.
  4. Simplify the radical, reduce the fraction by factoring the whole numerator, and state both solutions.
  5. Substitute both back into the original equation.

125. Pattern: pick the fastest method

Pattern

All four methods give the same answers. Choosing well is purely about how much arithmetic you sign up for.

what the equation looks likefastest methodwhy
no middle termsquare root propertyisolate the square and root it - two moves
already a squared binomialsquare root propertythe square is already built for you
factors easily over the integersfactoringreading two small numbers beats any formula
leading coefficient one, middle coefficient evencompleting the squarethe halving stays whole and the algebra is clean
anything else, or you are unsurequadratic formulait never fails and never needs a lucky guess
  1. Put the equation in standard form and read a, b, and c with their signs.
  2. Scan for a missing middle term or an obvious factorization - those are the shortcuts.
  3. If neither shortcut is there, compute the discriminant and use the quadratic formula.
  4. Simplify the radical, reduce the fraction by factoring the whole numerator, and state both solutions.
  5. Substitute both back into the original equation.

126. Fill in: why for Pattern: pick the fastest method

Comparison

Comparison matrix

From Pattern: pick the fastest method: refill the why column from what you know. The rest of the table is as it appeared.

what the equation looks likefastest methodwhy
no middle termsquare root propertyisolate the square and root it - two moves
already a squared binomialsquare root propertythe square is already built for you
factors easily over the integersfactoringreading two small numbers beats any formula
leading coefficient one, middle coefficient evencompleting the squarethe halving stays whole and the algebra is clean
anything else, or you are unsurequadratic formulait never fails and never needs a lucky guess

127. Choosing a method

Picture it

Animation

Shows: Choosing a method — a rendered Manim animation.

Rendered with Manim.

Takeaway: The formula always works and is rarely the fastest.

128. Rule out three: Check yourself: choosing the method

Elimination

Eliminate the wrong options

Which method is the most efficient choice for this equation?

3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.

  • A. Isolate the square and use the square root property
  • B. Substitute into the quadratic formula
  • C. Complete the square
  • D. Split the middle term and factor by grouping

Survives elimination: A

Why: There is no middle term, so two moves finish it: divide by 5 to get the square equal to 9, then take the root with plus-or-minus to get 3 and negative 3. Checking: 5 times 9 minus 45 is 0 for both.

129. Check yourself: choosing the method

Check

Every method would work here. Which one finishes in the fewest steps?

\[ 5x^2 - 45 = 0 \]

Check your understanding

Which method is the most efficient choice for this equation?

  • A. Isolate the square and use the square root property (correct)
  • B. Substitute into the quadratic formula
  • C. Complete the square
  • D. Split the middle term and factor by grouping

Answer: A

Why: There is no middle term, so two moves finish it: divide by 5 to get the square equal to 9, then take the root with plus-or-minus to get 3 and negative 3. Checking: 5 times 9 minus 45 is 0 for both.

Why B tempts people
The formula does work, but with b equal to zero it spends several lines of arithmetic rebuilding the square root property you could have used directly.
Why C tempts people
There is no middle coefficient to halve and square, so the completing step adds nothing and only lengthens the work.
Why D tempts people
Grouping needs four terms after splitting a middle term, and this equation has only two terms - there is nothing to split or group.

130. Applications come with a domain

Concept

The algebra will hand you two solutions. The situation decides how many of them are answers.

Reject the impossible root out loud, in a sentence. A rejected root is part of the answer, not a mistake you hide.

131. Break it if you can: Applications come with a domain

Counterexample

Discussion prompt

The algebra will hand you two solutions. The situation decides how many of them are answers.

That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.

Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.

Answer:

Reject the impossible root out loud, in a sentence. A rejected root is part of the answer, not a mistake you hide.

132. Plan first: Worked example: when does the ball land

Step zero

Discussion prompt

Worked example: when does the ball land — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.

Hint: It starts with: Translate the question into an equation

Answer:

  1. Translate the question into an equation
  2. Divide every term by negative 16
  3. Factor and apply the zero-product property
  4. Reject the negative time
  5. Verify the surviving solution in the original model

133. Worked example: when does the ball land

Worked example

A ball is thrown upward from the top of a sixty-four foot building at an initial speed of forty-eight feet per second. Its height in feet after t seconds is modeled by:

\[ h(t) = -16t^2 + 48t + 64 \]

Translate the question into an equation

Why: Hitting the ground means the height is zero, so set the model equal to zero.

\[ -16t^2 + 48t + 64 = 0 \]

Divide every term by negative 16

Why: This clears the big numbers and makes the leading coefficient positive, which makes factoring much easier. Zero divided by anything is still zero.

\[ t^2 - 3t - 4 = 0 \]

Factor and apply the zero-product property

Why: Negative four and one multiply to negative four and add to negative three.

\[ (t - 4)(t + 1) = 0 \;\Rightarrow\; t = 4 \quad \text{or} \quad t = -1 \]

Reject the negative time

Why: The model starts the clock at the moment of the throw, so a negative time describes an instant before the ball existed in this problem. Only one root survives.

\[ t = 4 \text{ seconds} \]

Verify the surviving solution in the original model

Why: Sixteen times sixteen is 256, forty-eight times four is 192, and negative 256 plus 192 plus 64 is exactly zero - the ball is at ground level at four seconds. The starting height also checks: at time zero the model gives 64 feet.

\[ h(4) = -16(16) + 48(4) + 64 = -256 + 192 + 64 = 0 \]

134. when does the ball land — line by line

Picture it

Animation

Shows: Each line of the worked example "when does the ball land", appearing one at a time.

The same working the example does, in the order a tutor would write it.

Takeaway: Sixteen times sixteen is 256, forty-eight times four is 192, and negative 256 plus 192 plus 64 is exactly zero - the ball is at ground level at four seconds. The starting height also checks: at time zero the model gives 64 feet.

135. What has to happen first: Worked example: a garden with a fixed area

Ranking

Put in order

Put the moves of Worked example: a garden with a fixed area into the order they have to happen.

  1. Define one variable and express the other in terms of it
  2. Write the area equation and expand it
  3. Subtract 54 to reach standard form
  4. Factor and solve
  5. Reject the negative width and report both dimensions
  6. Verify against the original wording

Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Naming the width first is the convention that keeps the length expression simple.

136. Worked example: a garden with a fixed area

Worked example

A rectangular garden is three feet longer than it is wide, and it covers fifty-four square feet. Find both dimensions.

Define one variable and express the other in terms of it

Why: Naming the width first is the convention that keeps the length expression simple. Everything else must be written using that one letter.

\[ \text{width} = w, \qquad \text{length} = w + 3 \]

Write the area equation and expand it

Why: Area of a rectangle is length times width, and expanding puts it in a form we can rearrange.

\[ w(w + 3) = 54 \;\Rightarrow\; w^2 + 3w = 54 \]

Subtract 54 to reach standard form

Why: The zero-product property needs zero alone on one side before any factoring can be used.

\[ w^2 + 3w - 54 = 0 \]

Factor and solve

Why: Nine and negative six multiply to negative fifty-four and add to three.

\[ (w + 9)(w - 6) = 0 \;\Rightarrow\; w = -9 \quad \text{or} \quad w = 6 \]

Reject the negative width and report both dimensions

Why: A garden cannot have a width of negative nine feet, so that root is discarded. The width is six feet, which makes the length nine feet.

\[ \text{width} = 6 \text{ ft}, \qquad \text{length} = 9 \text{ ft} \]

Verify against the original wording

Why: Nine is three more than six, so the length condition holds, and six times nine is fifty-four square feet, so the area condition holds too.

condition from the problemcheckholds?
length is 3 more than width9 = 6 + 3yes
area is 54 square feet6 times 9 = 54yes

137. Worked example: hitting a revenue target

Worked example

A shop sells a mug at p dollars each. At that price it sells a quantity given by two hundred minus five times the price, per week. Revenue is price times quantity. What price produces fifteen hundred dollars of weekly revenue?

\[ R(p) = p(200 - 5p) \]

Set the revenue model equal to the target

Why: The question fixes the output of the model, so the model becomes an equation in the price.

\[ p(200 - 5p) = 1500 \]

Expand and move everything to one side

Why: Distributing gives two hundred p minus five p squared, and standard form requires zero on the other side.

\[ 200p - 5p^2 = 1500 \;\Rightarrow\; 5p^2 - 200p + 1500 = 0 \]

Divide every term by 5

Why: Every coefficient shares a factor of five, and shrinking them makes the factoring obvious.

\[ p^2 - 40p + 300 = 0 \]

Factor and solve

Why: Negative ten and negative thirty multiply to three hundred and add to negative forty.

\[ (p - 10)(p - 30) = 0 \;\Rightarrow\; p = 10 \quad \text{or} \quad p = 30 \]

Keep both roots and interpret them

Why: Both prices are positive and both keep the quantity sold positive, so both are genuine business answers - a low price selling many mugs, or a high price selling few.

\[ 0 < p < 40 \]

Verify both prices in the original model

Why: At ten dollars the shop sells one hundred fifty mugs for fifteen hundred dollars. At thirty dollars it sells fifty mugs, also for fifteen hundred dollars.

pricequantity soldrevenue
10200 - 50 = 1501500
30200 - 150 = 501500

138. Worked example: a right triangle

Worked example

One leg of a right triangle is seven centimeters longer than the other, and the hypotenuse is thirteen centimeters. Find the legs.

Figure (svg): A right triangle with the shorter leg labeled x, the longer leg labeled x plus 7, and the hypotenuse labeled 13

Name the shorter leg first; the longer one is written from it.

Name the shorter leg and write the other side from it

Why: One letter has to carry the whole problem, so the longer leg is expressed as that letter plus seven.

\[ \text{legs: } x \text{ and } x + 7, \qquad \text{hypotenuse: } 13 \]

Apply the Pythagorean relationship

Why: In a right triangle the squares of the two legs add to the square of the hypotenuse.

\[ x^2 + (x + 7)^2 = 13^2 \]

Expand the squared binomial carefully

Why: The square of the quantity x plus seven is x squared plus fourteen x plus forty-nine - the middle term is the piece most often lost here.

\[ x^2 + x^2 + 14x + 49 = 169 \]

Combine and reduce to standard form, then divide by 2

Why: Two x squared plus fourteen x minus one hundred twenty equals zero, and every term shares a factor of two.

\[ 2x^2 + 14x - 120 = 0 \;\Rightarrow\; x^2 + 7x - 60 = 0 \]

Factor, solve, and reject the negative length

Why: Twelve and negative five multiply to negative sixty and add to seven. A leg of negative twelve centimeters is impossible, so only five survives.

\[ (x + 12)(x - 5) = 0 \;\Rightarrow\; x = 5 \text{ cm}, \quad x + 7 = 12 \text{ cm} \]

Verify the legs against the original triangle

Why: Twelve is seven more than five, and twenty-five plus one hundred forty-four is one hundred sixty-nine, which is thirteen squared. Both conditions hold.

\[ 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \]

139. Decode the notation: Worked example: a right triangle

Notation

Annotate

From Worked example: a right triangle — read this one piece at a time. What is each part doing?

On: \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \)

  • One letter has to carry the whole problem, so the longer leg is expressed as that letter plus seven.
  • In a right triangle the squares of the two legs add to the square of the hypotenuse.
  • The square of the quantity x plus seven is x squared plus fourteen x plus forty-nine - the middle term is the piece most often lost here.

140. Check yourself: an applied quadratic

Check

A rectangular patio is four feet longer than it is wide and covers ninety-six square feet. Set it up, solve it, and then read the question again before choosing.

Check your understanding

What is the width of the patio?

  • A. 8 feet (correct)
  • B. -12 feet
  • C. Both 8 feet and -12 feet
  • D. 12 feet

Answer: A

Why: Letting the width be w, the area equation is w times the quantity w plus 4 equals 96, which becomes w squared plus 4w minus 96 equals 0 and factors as w plus 12 times w minus 8. Only the positive root is a width, and 8 times 12 is 96.

Why B tempts people
Reported the rejected root; a physical width cannot be a negative number of feet, so this solution of the equation is not a solution of the problem.
Why C tempts people
Gave both algebraic roots without applying the domain check that the situation forces on the answer.
Why D tempts people
Solved correctly but answered with the length instead of the width; 12 is the value of w plus 4, not of w.

141. Rebuild the recipe: Pattern: the whole-unit playbook

Ranking

Put in order

These are the steps of Pattern: the whole-unit playbook, scrambled. Put them back in order before the next slide shows you.

  1. Standard form first. Expand, clear fractions, and move everything to one side so zero is alone.
  2. Read a, b, and c with their signs. Write them on their own line.
  3. Look for a shortcut. No middle term means the square root property; an easy factorization means factoring.
  4. Otherwise compute the discriminant, then use the quadratic formula with every value in parentheses.

Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.

142. Pattern: the whole-unit playbook

Pattern

  1. Standard form first. Expand, clear fractions, and move everything to one side so zero is alone.
  2. Read a, b, and c with their signs. Write them on their own line.
  3. Look for a shortcut. No middle term means the square root property; an easy factorization means factoring.
  4. Otherwise compute the discriminant, then use the quadratic formula with every value in parentheses.
  1. Simplify the radical, then reduce by factoring the entire numerator - never cancel into a single term.
  2. Keep the plus-or-minus all the way to the end, and state both solutions.
  3. Verify by substituting each solution into the original equation.
  4. In a word problem, reject the roots the context forbids and answer the question that was actually asked, with units.

Steps one, six, and seven are where nearly all the lost points live. They are also the cheapest ones to protect.

143. Where this shows up: Solving Quadratic Equations

Real world

Discussion prompt

Outside this lesson: where does Solving Quadratic Equations actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: the whole-unit playbook is doing the work in it.

Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.

Answer:

That deck covers every way to solve a quadratic equation, and how to pick the fastest one. It starts with standard form and the zero-product property, then works through factoring, the square root property, completing the square, the quadratic formula derived from it, and what the discriminant says about the number and type of solutions, before finishing with projectile, area, revenue, and Pythagorean applications. It targets the four errors that cost the most points: setting factors equal to a constant instead of zero, losing half the solutions by dropping the plus-or-minus, dropping the sign of a negative coefficient inside the quadratic formula, and dividing only one term by the leading coefficient while completing the square.

144. Squaring can invent solutions

Picture it

Animation

Shows: Squaring can invent solutions — a rendered Manim animation.

Rendered with Manim.

Takeaway: Squaring is not reversible, which is why checking is mandatory.

145. Connect it up: Solving Quadratic Equations

Connect it up

Draw it

One page, no notation unless you need it: draw how these connect — What Makes an Equation Quadratic · Factoring and the Zero-Product Property · The Square Root Property · Completing the Square · The Quadratic Formula · The Discriminant. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.

146. What you can do now

Recap

A quadratic equation is a question about where a parabola meets zero - and you now own four different ways to answer it.

methoduse it whenthe move that defines it
factoringthe trinomial factors over the integerszero-product property, after standard form
square root propertythere is no middle termroot both sides, keeping plus-or-minus
completing the squareyou want vertex form, or b is evenadd the square of half the middle coefficient to both sides
quadratic formulaalways, and whenever you are unsuresubstitute a, b, and c in parentheses

Next up: radical, rational, and quadratic-form equations, where the same solving instincts run into extraneous solutions and every candidate has to be checked in the original equation.

Sources

  1. OpenStax College Algebra 2e
  2. All algebra, solutions, and numeric results re-derived and verified by hand. — Verified 2026-07-31.

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