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
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.
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.
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.
Section
Part 1
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.
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.
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.
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:
\[ x^2 - 9 = 0 \qquad\text{and}\qquad 3x^2 + 12x = 0 \]
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.
Picture it
Figure (svg): A U-shaped parabola crossing a horizontal axis at two marked points
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.
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 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.
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.
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.
Ranking
Put in order
Put the moves of Worked example: rewrite in standard form into the order they have to happen.
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.
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.
\[ 3x^2 - 11x - 4 = 0 \;\Longleftrightarrow\; 3x^2 - 6x = 5x + 4 \;\Longleftrightarrow\; 3x(x-2) = 5x + 4 \]
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.
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.
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.
Pattern
Do this before deciding which solving method to use. The coefficients you read here are what the decision depends on.
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?
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.
Section
Part 2
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.
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.
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.
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.
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:
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.
| candidate | left side computed | equals 0? |
|---|---|---|
| x = -2 | 4 - 10 + 6 | yes |
| x = -3 | 9 - 15 + 6 | yes |
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.
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.
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.
| candidate | left side | should be 8 |
|---|---|---|
| x = 8 | 64 - 16 = 48 | no |
| x = 10 | 100 - 20 = 80 | no |
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.
| solution | left side | should be 8 |
|---|---|---|
| x = 4 | 16 - 8 = 8 | yes |
| x = -2 | 4 + 4 = 8 | yes |
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.
| candidate | left side | should be 8 |
|---|---|---|
| x = 8 | 64 - 16 = 48 | no |
| x = 10 | 100 - 20 = 80 | no |
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.
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.
| candidate | computation | result |
|---|---|---|
| x = 1/2 | 0.5 - 3.5 + 3 | 0 |
| x = 3 | 18 - 21 + 3 | 0 |
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.
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.
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.
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.
| candidate | left side | right side |
|---|---|---|
| x = 0 | 0 | 0 |
| x = 4 | 48 | 48 |
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.
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.
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.
| solution | found? |
|---|---|
| x = 4 | yes |
| x = 0 | lost |
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.
| solution | found? |
|---|---|
| x = 4 | yes |
| x = 0 | yes |
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.
| solution | found? |
|---|---|
| x = 4 | yes |
| x = 0 | lost |
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.
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.
\[ 5^2 - 10(5) + 25 = 25 - 50 + 25 = 0 \]
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.
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?
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.
Section
Part 3
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.
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.
Picture it
Figure (svg): A number line with points at negative five and five, both marked as squaring to twenty-five
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.
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
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.
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.
\[ (x - 4)^2 = 18 \;\Rightarrow\; x - 4 = \pm\sqrt{18} \]
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.
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.
| candidate | left side computed | equals 0? |
|---|---|---|
| x = 5 | 3(25) - 75 = 0 | yes |
| x = -5 | 3(25) - 75 = 0 | yes |
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.
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.
| value | value squared | solves the equation? |
|---|---|---|
| 7 | 49 | yes |
| -7 | 49 | yes, but was dropped |
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.
| solution | check | result |
|---|---|---|
| x = 7 | 7 squared | 49 |
| x = -7 | (-7) squared | 49 |
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.
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:
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.
\[ \left(4 + 3\sqrt{2} - 4\right)^2 = \left(3\sqrt{2}\right)^2 = 9 \cdot 2 = 18 \]
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.
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.
\[ 2(4i)^2 + 32 = 2(16)(-1) + 32 = -32 + 32 = 0 \]
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.
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?
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.
Section
Part 4
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
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.
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
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.
Concept
Every perfect-square trinomial has the same fingerprint: the constant is the square of half the middle coefficient.
\[ x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2 \]
| expression | half the middle coefficient | its square | factors as |
|---|---|---|---|
| x squared plus 6x | 3 | 9 | (x + 3) squared |
| x squared minus 8x | -4 | 16 | (x - 4) squared |
| x squared plus 5x | 5/2 | 25/4 | (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.
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.
| expression | half the middle coefficient | its square | factors as |
|---|---|---|---|
| x squared plus 6x | 3 | 9 | (x + 3) squared |
| x squared minus 8x | -4 | 16 | (x - 4) squared |
| x squared plus 5x | 5/2 | 25/4 | (x + 5/2) squared |
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.
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.
| candidate | left side computed | equals 0? |
|---|---|---|
| x = -1 | 1 - 6 + 5 | yes |
| x = -5 | 25 - 30 + 5 | yes |
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.
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.
\[ (4 + \sqrt{13})^2 - 8(4 + \sqrt{13}) + 3 = 29 + 8\sqrt{13} - 32 - 8\sqrt{13} + 3 = 0 \]
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.
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.
Ranking
Put in order
Put the moves of Worked example: divide through first into the order they have to happen.
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.
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.
| candidate | computation | result |
|---|---|---|
| x = 1 | 2(1) + 8(1) - 10 | 0 |
| x = -5 | 2(25) + 8(-5) - 10 | 0 |
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.
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.
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.
| candidate | original left side | equals 0? |
|---|---|---|
| about 1.14 | about 2.6 | no |
| about -13.14 | about 345.4 | no |
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.
| solution | original left side | equals 0? |
|---|---|---|
| x = 1 | 3 + 12 - 15 | yes |
| x = -5 | 75 - 60 - 15 | yes |
Notation
Annotate
From Trap: dividing only the squared term — read this one piece at a time. What is each part doing?
On: \( 3x^2 + 12x - 15 = 0 \)
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.
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.
\[ 3(x-2)^2 - 10 = 3\left(\frac{10}{3}\right) - 10 = 0, \qquad 3(x-2)^2 - 10 = 3x^2 - 12x + 2 \]
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.
Pattern
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.
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.
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?
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.
Section
Part 5
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.
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.
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.
\[ x = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2} = -2 \;\text{ or }\; -3 \]
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.
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.
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:
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.
| candidate | computation | result |
|---|---|---|
| x = 1/2 | 0.5 + 2.5 - 3 | 0 |
| x = -3 | 18 - 15 - 3 | 0 |
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.
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.
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.
| candidate | computation | result |
|---|---|---|
| x = 2 | 3(4) - 5(2) - 2 | 0 |
| x = -1/3 | 1/3 + 5/3 - 2 | 0 |
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.
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.
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.
| candidate | original left side | equals 0? |
|---|---|---|
| about -1.586 | about 19.0 | no |
| about -4.414 | about 52.97 | no |
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.
\[ (3 + \sqrt{2})^2 - 6(3 + \sqrt{2}) + 7 = 11 + 6\sqrt{2} - 18 - 6\sqrt{2} + 7 = 0 \]
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.
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.
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.
\[ 2x^2 - 4x - 3 = \left(7 + 2\sqrt{10}\right) - \left(4 + 2\sqrt{10}\right) - 3 = 0 \]
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.
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.
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.
| candidate | original left side | equals 0? |
|---|---|---|
| about 7.325 | 75 | no |
| about -5.325 | 75 | no |
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.
| candidate | original left side | equals 0? |
|---|---|---|
| about 2.581 | 0 | yes |
| about -0.581 | 0 | yes |
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.
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?
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.
Section
Part 6
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.
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.
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.
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.
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
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.
Concept
| discriminant | number and type of solutions | what the parabola does |
|---|---|---|
| positive | two different real solutions | crosses the horizontal axis twice |
| zero | one repeated real solution | touches the axis at exactly one point |
| negative | two complex conjugate solutions | never 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
One more free fact: when the discriminant is a perfect square, the radical disappears, the answers are rational, and the equation would have factored.
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.
| discriminant | number and type of solutions | what the parabola does |
|---|---|---|
| positive | two different real solutions | crosses the horizontal axis twice |
| zero | one repeated real solution | touches the axis at exactly one point |
| negative | two complex conjugate solutions | never reaches the axis |
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.
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.
\[ 4\left(\tfrac{9}{4}\right) - 12\left(\tfrac{3}{2}\right) + 9 = 9 - 18 + 9 = 0 \]
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.
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.
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.
\[ (-1+2i)^2 + 2(-1+2i) + 5 = (-3-4i) + (-2+4i) + 5 = 0 \]
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.
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.
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.
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?
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.
Section
Part 7
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:
Pattern
All four methods give the same answers. Choosing well is purely about how much arithmetic you sign up for.
| what the equation looks like | fastest method | why |
|---|---|---|
| no middle term | square root property | isolate the square and root it - two moves |
| already a squared binomial | square root property | the square is already built for you |
| factors easily over the integers | factoring | reading two small numbers beats any formula |
| leading coefficient one, middle coefficient even | completing the square | the halving stays whole and the algebra is clean |
| anything else, or you are unsure | quadratic formula | it never fails and never needs a lucky guess |
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 like | fastest method | why |
|---|---|---|
| no middle term | square root property | isolate the square and root it - two moves |
| already a squared binomial | square root property | the square is already built for you |
| factors easily over the integers | factoring | reading two small numbers beats any formula |
| leading coefficient one, middle coefficient even | completing the square | the halving stays whole and the algebra is clean |
| anything else, or you are unsure | quadratic formula | it never fails and never needs a lucky guess |
Picture it
Animation
Shows: Choosing a method — a rendered Manim animation.
Rendered with Manim.
Takeaway: The formula always works and is rarely the fastest.
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.
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.
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?
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.
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.
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.
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:
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.
\[ h(4) = -16(16) + 48(4) + 64 = -256 + 192 + 64 = 0 \]
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.
Ranking
Put in order
Put the moves of Worked example: a garden with a fixed area into the order they have to happen.
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.
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 problem | check | holds? |
|---|---|---|
| length is 3 more than width | 9 = 6 + 3 | yes |
| area is 54 square feet | 6 times 9 = 54 | yes |
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.
| price | quantity sold | revenue |
|---|---|---|
| 10 | 200 - 50 = 150 | 1500 |
| 30 | 200 - 150 = 50 | 1500 |
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 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.
\[ 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \]
Notation
Annotate
From Worked example: a right triangle — read this one piece at a time. What is each part doing?
On: \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \)
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?
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.
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.
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.
Pattern
Steps one, six, and seven are where nearly all the lost points live. They are also the cheapest ones to protect.
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.
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.
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.
Recap
A quadratic equation is a question about where a parabola meets zero - and you now own four different ways to answer it.
| method | use it when | the move that defines it |
|---|---|---|
| factoring | the trinomial factors over the integers | zero-product property, after standard form |
| square root property | there is no middle term | root both sides, keeping plus-or-minus |
| completing the square | you want vertex form, or b is even | add the square of half the middle coefficient to both sides |
| quadratic formula | always, and whenever you are unsure | substitute 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.
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