Solving a quadratic equation with no linear term by isolating the squared variable and taking square roots. Includes the standard form of a quadratic equation and the leading coefficient, the three cases for how many solutions an equation has, rewriting before taking roots, and using the falling object model to answer a question about a dropped object.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 9 — Quadratic Equations and Functions
Solving Quadratic Equations by Finding Square Roots
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 505-510 — the lesson these objectives are drawn from
Warm-up
Lesson 9.1 evaluated square roots. This lesson uses them to undo a squaring inside an equation.
Discussion prompt
Name every number that makes x squared equal 36. Then say how you would find them without guessing.
Hint: Two numbers work, not one.
Answer:
\[ x^2 = 36 \;\Longrightarrow\; x = \pm\sqrt{36} = \pm 6 \]
Taking square roots of both sides undoes the squaring, and the plus-or-minus sign is needed because six and negative six both square to thirty-six. Solving by inspection works for small perfect squares and fails immediately for anything else, which is why the method matters.
Concept
A quadratic equation is one that can be written as a times x squared plus b times x plus c equals nought, with a not nought. When b is nought, isolating the squared term and taking square roots of both sides solves it.
quadratic equation — An equation that can be written in the standard form a x squared plus b x plus c equals nought, where a is not nought. The coefficient a is called the leading coefficient.
The plus-or-minus sign of Lesson 9.1 becomes a second solution here.
Figure (svg): The standard form of a quadratic equation with its parts labelled
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 505-505
Section
Section 1
Concept
If the squared variable is already alone, taking square roots of both sides finishes the problem. Write the solutions as integers when the number is a perfect square and as radical expressions otherwise.
\[ x^2 = d \;\Longrightarrow\; x = \pm\sqrt{d} \]
Squaring and taking a square root are inverse operations.
Figure (svg): Squaring and taking a square root shown as inverse machines
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 505-505 — Example 1, Solve Quadratic Equations, and its Study Tip on inverse operations
Picture it
Squaring loses the sign.
Figure (svg): Squaring and taking a square root shown as inverse machines
Because squaring sends four and negative four to the same place, undoing it cannot know which one to return to. The plus-or-minus sign is how the notation admits that.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } x^2 = 4 \text{ and } n^2 = 5. \]
Take roots of the first
Why: The variable is already alone.
\[ x = \pm\sqrt{4} \]
Evaluate
Why: Four is a perfect square.
\[ x = \pm 2 \]
Take roots of the second
Why: Five is not a perfect square.
\[ n = \pm\sqrt{5} \]
Leave it exact
Why: A radical expression is the answer.
\[ n = \pm\sqrt{5} \]
Figure (svg): Squaring and taking a square root shown as inverse machines
\[ x = \pm 2, \qquad n = \pm\sqrt{5} \]
Verify: square each root
Why: Two squared and negative two squared are both four, and the root of five squared is five whichever sign it carries. Squaring the answer is the check that works every time here, because squaring is precisely what the equation does.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 505-505
Matching
Exact where the number allows.
Match the pairs
Why: Two of these radicands are perfect squares and two are not, which is the only difference between the tidy answers and the radical ones. Every one of the four has two solutions.
Worked example
Guided Practice 1 to 4, exact where possible.
\[ \text{Solve } x^2 = 81, \; y^2 = 11, \; n^2 = 25 \text{ and } x^2 = 10. \]
Take the first two
Why: Eighty-one is a perfect square; eleven is not.
\[ \pm 9, \; \pm\sqrt{11} \]
Take the third
Why: Twenty-five is five squared.
\[ \pm 5 \]
Take the fourth
Why: Ten is not a perfect square.
\[ \pm\sqrt{10} \]
Compare the forms
Why: Two are integers and two stay as radicals.
Figure (svg): The solution to Worked example four more of the same shape shown as a ladder of expressions, one row per algebraic move
\[ \pm 9, \quad \pm\sqrt{11}, \quad \pm 5, \quad \pm\sqrt{10} \]
Verify: check one of the radical answers
Why: The root of eleven squared is exactly eleven, so it satisfies the equation with nothing lost. Rounding it to 3.32 and squaring gives 11.0224, which does not satisfy the equation, and that is why the radical form is the correct answer rather than a decimal.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 505-505
Trap
\[ x^2 = 49 \;\Longrightarrow\; x = 7 \]
Take the square root of both sides
Why: The radical gives seven, so seven is the answer.
Negative seven also squares to forty-nine, so it solves the equation too. Half the answer has been lost, and on a graph it is the whole left-hand solution.
\[ x^2 = 49 \;\Longrightarrow\; x = \pm\sqrt{49} = \pm 7 \]
Write the plus-or-minus sign when taking roots to solve
Why: Solving asks for every value that works.
The bare radical means the positive root, which is exactly why the plus-or-minus has to be written in by hand at this step.
Faded example
The plus-or-minus goes in at this step.
Fill in the blanks
x^2 = 36 \;\Longrightarrow\; x = @pm\sqrt@pm 6 = ___
Why: Solving an equation asks for every value that works, so both roots are wanted. The bare radical would give only six, which is why the symbol has to be written in deliberately.
Elimination
Solving x squared equals 100.
Eliminate the wrong options
Which is the full solution set?
Survives elimination: A
Why: The two solutions are always symmetric about nought when the equation has this shape. Option C is worth naming: halving is not the inverse of squaring, and confusing the two is a persistent early error.
Socratic
Undoing an addition gives one.
Discussion prompt
Explain why taking a square root of both sides produces two solutions when subtracting from both sides produces one. Then say what property of squaring is responsible.
Hint: Ask whether two different inputs can share an output.
Answer:
Adding three to a number and adding three to a different number always give different results, so undoing the addition can only lead back to one place. Squaring is not like that: four and negative four are different inputs with the same output, so the operation loses information about the sign.
Undoing an operation that has lost information cannot recover what was lost, so it must offer every possibility instead. That is what the plus-or-minus sign is doing, and it is why quadratic equations generally have two solutions while linear ones have exactly one.
Section
Section 2
Concept
For an equation of the form x squared equals d, the number of real solutions depends only on the sign of d. Positive gives two, nought gives one, and negative gives none.
The square of a real number is never negative.
Figure (svg): The three cases for the number of solutions
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-506 — Example 2 and the Summary box on solving x squared equals d
Picture it
Decided before any work.
Figure (svg): The three cases for the number of solutions
This is the same three-case split as Lesson 9.1's count of square roots, now phrased as a statement about solutions of an equation.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } x^2 = 0 \text{ and } y^2 = -1. \]
Take roots of the first
Why: Only nought squares to nought.
\[ x = 0 \]
Count its solutions
Why: The two roots coincide.
Look at the second
Why: The right side is negative.
\[ y ^{2} = -1 \]
Conclude
Why: No real number squares to a negative.
Figure (svg): The three cases for the number of solutions
\[ x = 0; \qquad y^2 = -1 \text{ has no real solution} \]
Verify: try a few values in the second
Why: One squared is one, negative one squared is one, and nought squared is nought — all positive or nought. Nothing at all is available to square to negative one, which is what having no real solution means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-506
Sorting
Read the sign of the right side.
Sort into buckets
Sort each equation by its number of real solutions.
Not one of these needed solving to be sorted. Whether the answers are whole numbers never came into it either — only the sign did.
Worked example
Reading the sign of d first.
\[ \text{How many real solutions have } x^2 = 7, \; x^2 = 0 \text{ and } x^2 = -7? \]
Look at the first
Why: Seven is positive.
Look at the second
Why: The right side is nought.
Look at the third
Why: Negative seven is negative.
Give the first one exactly
Why: Seven is not a perfect square.
\[ \pm\sqrt{7} \]
Figure (svg): The solution to Worked example sort three equations without solving shown as a ladder of expressions, one row per algebraic move
\[ \pm\sqrt{7}; \quad 0; \quad \text{no real solution} \]
Verify: check that the count came before the work
Why: Only the first equation needed any computation, and even there the sign of d was enough to know two answers were coming. Reading the sign first turns a third of these problems into one-line answers.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-506
Trap
\[ y^2 = -1 \;\Longrightarrow\; y = \pm\sqrt{-1} = \pm 1 \]
Take roots of both sides and drop the awkward minus
Why: The minus sign looked like a stray.
Checking undoes it at once: one squared is positive one, not negative one, and the same for negative one. The equation has no real solution and no amount of manipulation produces one.
The equation y squared equals negative one has no real solution.
Read the sign of the right side before taking any roots
Why: A negative there ends the problem.
Saying so is a complete and correct answer, not an admission of defeat.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Sign of d | How many solutions | What they are |
|---|---|---|
| positive | two | plus and minus the root of d |
| zero | one | zero |
| negative | none | no real number squares to a negative |
The middle row is the boundary between the other two, which is why it has an odd number of solutions. As d shrinks to nought the two solutions meet, and past nought they vanish.
Prediction
Watch the two solutions of x squared equals d.
Predict first
As d decreases towards nought, what do the two solutions do?
Correct: They move towards each other and meet at nought.
\[ d = 4: \; \pm 2 \qquad d = 1: \; \pm 1 \qquad d = 0.01: \; \pm 0.1 \]
Why: The solutions are plus and minus the root of d, so both shrink in size as d does — the solutions of x squared equals four are two apart from nought, and those of x squared equals one are only one apart. They collide at nought, which is exactly why that case has a single solution, and past it they have nowhere real to go.
Socratic
Two and none feel more natural.
Discussion prompt
Explain why exactly one solution happens only when d is nought. Then say what that has to do with the two solutions being symmetric.
Hint: Ask where the two solutions sit relative to each other.
Answer:
The solutions always come as a number and its negative, so they are mirror images across nought and are therefore different from each other — unless the number is nought itself, which is its own negative. That single exception is the only way the pair can collapse into one value.
So the symmetry forces the count to be two or nought in every other case, and one solution can only occur at the exact centre. Lesson 9.5 shows this graphically: the curve either crosses the axis twice, touches it once at its lowest point, or misses it entirely.
Section
Section 3
Concept
When the equation is not already in the form x squared equals d, use inverse operations to isolate the squared term. Only then take square roots of both sides.
The first two steps are ordinary Chapter 3 equation solving.
Figure (svg): Isolating the squared term before taking roots
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-506 — Example 3, Rewrite Before Finding Square Roots
Picture it
Constant, coefficient, root.
Figure (svg): Isolating the squared term before taking roots
Nothing in the first two steps is new. Recognising that a quadratic equation of this shape yields to familiar moves is most of what makes it easy.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } 3x^2 - 48 = 0. \]
Move the constant
Why: Add forty-eight to each side.
\[ 3 x ^{2} = 48 \]
Divide by the coefficient
Why: Each side by three.
\[ x ^{2} = 16 \]
Take square roots
Why: Both signs.
\[ x = \pm\sqrt{16} \]
Evaluate
Why: Sixteen is a perfect square.
\[ x = \pm 4 \]
Figure (svg): Isolating the squared term before taking roots
\[ x = \pm 4 \]
Verify: check both in the original equation
Why: Three times sixteen is forty-eight, and forty-eight minus forty-eight is nought — and negative four squared is also sixteen, so it gives the same line. Both values satisfy the original equation, which confirms there really are two solutions.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-506
Faded example
Two ordinary steps before the new one.
Fill in the blanks
2x^2 - 72 = 0 \;\to\; 2x^2 = 72 \;\to\; x^2 = 36 \;\to\; x = @pm 6
Why: Dividing by the leading coefficient has to happen before the root is taken. Doing it afterwards, or not at all, changes the answer by a factor that the check will catch.
Worked example
Guided Practice 5 to 7, including one with the terms swapped.
\[ \text{Solve } x^2 - 1 = 0, \; 2x^2 - 72 = 0 \text{ and } 27 - 3y^2 = 0. \]
Do the first
Why: Add one to each side.
\[ x^2 = 1 \Rightarrow x = \pm 1 \]
Do the second
Why: Add seventy-two, then divide by two.
\[ x^2 = 36 \Rightarrow x = \pm 6 \]
Read the third carefully
Why: The squared term is being subtracted.
\[ 27 = 3 y ^{2} \]
Finish the third
Why: Divide by three, then take roots.
\[ y^2 = 9 \Rightarrow y = \pm 3 \]
Figure (svg): The solution to Worked example three more rewrites shown as a ladder of expressions, one row per algebraic move
\[ \pm 1, \quad \pm 6, \quad \pm 3 \]
Verify: check the third one, where the order was unusual
Why: Twenty-seven minus three times nine is twenty-seven minus twenty-seven, which is nought, and the same holds for negative three. Moving the squared term to the other side rather than the constant works equally well and is often tidier when the squared term is negative.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-506
Error analysis
The student was solving three x squared minus forty-eight equals nought.
Annotate
On: \( \begin{aligned} 3x^2 - 48 &= 0 \\ 3x^2 &= 48 \\ x &= \pm\sqrt{48} \\ x &\approx \pm 6.93 \end{aligned} \)
The rule is that the squared term must be completely alone before the root is taken — no coefficient in front and nothing added to it. Taking a root of both sides of three x squared equals forty-eight would actually give the root of three times x, which is not a step anyone wants.
Sorting
For each equation, name the first move.
Sort into buckets
Sort each equation by what has to be done to it first.
Only two of the six were ready to go. The habit worth building is to check that the squared term is genuinely alone before reaching for the radical.
Elimination
Solving 5x squared equals 45.
Eliminate the wrong options
What is the correct first move?
Survives elimination: A
Why: Dividing by five gives x squared equals nine, and the roots follow. Option C is the more common error, treating a multiplier as though it were a term, and it comes from Chapter 3 habits applied without checking.
Socratic
The order of the steps is not free.
Discussion prompt
Explain why dividing by the leading coefficient has to happen before the square root and not after. Then say what taking the root of three x squared would actually give.
Hint: Ask what the root of a product does.
Answer:
A square root applies to the whole expression underneath it, so taking the root of three x squared gives the root of three times x, not three times x. The coefficient does not simply pass through the operation unchanged, which is why it has to be cleared away while the equation is still in ordinary form.
Removing it afterwards would mean dividing by the root of three rather than by three, which is a harder and easier-to-botch step. Dividing first keeps every number in the problem rational until the very last line, and that is worth doing whenever the choice is available.
Section
Section 4
Concept
When an object is dropped, its height h after t seconds is approximated by negative sixteen t squared plus s, where s is the height it was dropped from. Height is in feet and time in seconds, and air resistance is ignored.
\[ h = -16t^2 + s \]
A dropped object keeps speeding up, which the squared term produces.
Figure (svg): The height of a dropped egg against time
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 507-507 — the Falling Object Model and Example 4, Write a Falling Object Model
Picture it
Slow at first, fast at the end.
Figure (svg): The height of a dropped egg against time
The model has no x term, so it is exactly the shape this lesson can solve. That is not a coincidence — the section is built around it.
Worked example
This is Example 4 from the textbook.
\[ \text{An egg container is dropped from } 32 \text{ feet. Write its falling object model.} \]
Recall the model
Why: The general falling object equation.
\[ h = -16 t ^{2} + s \]
Identify the initial height
Why: The height it was dropped from.
\[ s = 32 \]
Substitute
Why: Put thirty-two in place of s.
\[ h = -16 t ^{2} + 32 \]
Check at time nought
Why: No time has passed yet.
\[ h = 32 \]
Figure (svg): The height of a dropped egg against time
\[ h = -16t^2 + 32 \]
Verify: test the model at the start
Why: At t equal to nought the squared term vanishes and the height is thirty-two, which is where the egg began. Any falling model should return the starting height at time nought, and it is a one-second check.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 507-507
Faded example
Landing means the height is nought.
Fill in the blanks
0 = -16t^2 + 32 \;\to\; -32 = -16t^2 \;\to\; t^2 = 2 \;\to\; t \approx 1.4
Why: Dividing two negatives gives a positive, so the squared term ends up equal to two. The negative root is dropped afterwards because a negative time has no meaning for a falling object.
Worked example
This is Example 5 from the textbook.
\[ \text{Using } h = -16t^2 + 32, \text{ find when the container reaches the ground.} \]
Set the height to nought
Why: Ground level.
\[ 0 = -16 t ^{2} + 32 \]
Move the constant
Why: Subtract thirty-two from each side.
\[ -32 = -16 t ^{2} \]
Divide
Why: Each side by negative sixteen.
\[ 2 = t ^{2} \]
Take roots and choose
Why: Time cannot be negative.
\[ t \approx 1.4 \]
Figure (svg): Two algebraic solutions, one of which the situation rejects
\[ t = \sqrt{2} \approx 1.4 \text{ seconds} \]
Verify: put the time back into the model
Why: Sixteen times two is thirty-two, so negative thirty-two plus thirty-two is nought — the egg is exactly at ground level. Substituting the answer back into the original model is the check that catches both algebra slips and a misread starting height.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 507-507
Trap
\[ t = \pm\sqrt{2} \approx \pm 1.4 \]
Report both solutions, as the algebra gives them
Why: Every other equation in this lesson had two answers.
A time of negative one and a half seconds would be before the egg was dropped, when the model does not apply. The equation does not know that; the person solving it has to.
\[ t = \sqrt{2} \approx 1.4 \text{ seconds} \]
Discard the solution that the situation rules out
Why: Time, length and count are never negative.
Say why it was discarded rather than silently dropping it; that is what distinguishes a modelling answer from an algebraic one.
Prediction
The same egg, dropped from 64 feet instead of 32.
Predict first
How does the falling time change?
Correct: It grows by a factor of about 1.4, not by a factor of 2.
\[ s = 32: \; t = \sqrt{2} \approx 1.4 \qquad s = 64: \; t = 2 \]
Why: From sixty-four feet the equation becomes t squared equals four, so the time is two seconds against the original root of two, about 1.41 seconds. Doubling the height multiplies the time by the root of two rather than by two, because the height depends on the square of the time. That is the same reasoning that makes a four-times-higher drop take exactly twice as long.
Elimination
Using h equals negative sixteen t squared plus thirty-two.
Eliminate the wrong options
What should be substituted?
Survives elimination: A
Why: Reaching the ground means the height is nought, so nought is what h becomes. Options B and C are worth comparing: they are the same instant read in two directions, and neither answers the question asked.
Socratic
The object was dropped, not thrown.
Discussion prompt
Explain what the missing linear term means physically. Then say what would change if the egg were thrown downwards instead of released.
Hint: Ask about the speed at the moment of release.
Answer:
A dropped object starts at rest, so at the first instant its height is not changing at all — the curve leaves the starting height horizontally. A linear term would represent an initial upward or downward speed, and dropping means that speed is nought, which is exactly why the term is absent.
Thrown downwards, the model would gain a negative linear term and the egg would land sooner; thrown upwards, the term would be positive and the egg would rise before falling. Both of those produce equations with all three terms, which square roots alone cannot solve — that is the job of Lessons 9.6 and 12.5.
Section
Section 5
Concept
Substituting each solution into the original equation confirms the algebra. Deciding which solutions to report is a separate question, answered by the situation rather than by the equation.
Times, lengths and counts are never negative.
Figure (svg): Checking two solutions in the original equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-507 — the check in Example 3 and the Study Tip in Example 5 on ignoring the negative root
Picture it
Squaring erases the sign.
Figure (svg): Checking two solutions in the original equation
The two checks are identical after the squaring step, which is the algebraic reason the solutions arrive as a symmetric pair.
Worked example
The check from Example 3, written out.
\[ \text{Verify that } 4 \text{ and } -4 \text{ both solve } 3x^2 - 48 = 0. \]
Substitute four
Why: Square it first.
\[ 3(4) ^{2} - 48 \]
Simplify
Why: Three times sixteen is forty-eight.
\[ 48 - 48 = 0 \]
Substitute negative four
Why: Squaring removes the sign.
\[ 3(-4) ^{2} - 48 \]
Simplify
Why: The same line as before.
\[ 48 - 48 = 0 \]
Figure (svg): Checking two solutions in the original equation
\[ 3(4)^2 - 48 = 0 \quad \text{and} \quad 3(-4)^2 - 48 = 0 \]
Verify: notice why they agree
Why: Once each value is squared the sign is gone, so the two checks become the same arithmetic. That is worth seeing once, because it explains why checking the second solution of this kind of equation almost never turns up a surprise.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 506-506
Hypothesis
Several lines are available.
Predict first
Substituting a solution into which line gives a meaningful check?
Correct: The original equation, as it was first given.
This is the same reasoning as checking a word problem against the words rather than against your own equation.
Why: Every rewritten line was produced by the same work that might contain the mistake, so checking against one of them can confirm an error rather than catch it. Only the original statement is independent of the solving, which is what makes it the right thing to test against. The lines are equivalent when the work is correct, and the point of a check is that you do not yet know whether it is.
Worked example
One equation asked twice.
\[ \text{Solve } t^2 = 2 \text{ as an equation, then as a landing time in seconds.} \]
Solve as an equation
Why: Both roots are wanted.
\[ t = \pm\sqrt{2} \]
Approximate
Why: To one decimal place.
\[ \pm 1.4 \]
Read as a time
Why: Negative seconds have no meaning here.
Report the time
Why: One value, with units.
\[ \text{about } 1.4\text{ seconds} \]
Figure (svg): Two algebraic solutions, one of which the situation rejects
\[ t = \pm\sqrt{2}; \qquad t \approx 1.4 \text{ seconds} \]
Verify: say what changed between the two answers
Why: The algebra did not change at all — only the question did. The second answer discards a mathematically valid solution on the grounds of what it would mean, which is a judgement the equation cannot make for you.
Trap
\[ 3x^2 - 48 = 0 \;\to\; x^2 = 16 \;\to\; x = \pm 4 \]
Check by substituting four into x squared equals sixteen
Why: It is the simplest of the lines, so it is the easiest to check.
That only confirms the last step. If the division by three had gone wrong, the rewritten equation would be wrong in the same way and the check would pass anyway.
\[ 3(4)^2 - 48 = 48 - 48 = 0 \;\checkmark \]
Substitute into the equation as it was originally given
Why: That is the statement the answer has to satisfy.
A check is only worth doing against something that could not have inherited the error.
Sorting
The algebra gives two answers each time.
Sort into buckets
Sort each question by how many of the two solutions should be reported.
The split is not about the mathematics, which is identical in every row, but about what the answer is being used for. Saying why a solution was rejected is part of the answer.
Faded example
Squaring removes the sign.
Fill in the blanks
3(-4)^2 - 48 = 3(16) - 48 = 0
Why: The brackets matter: negative four squared is positive sixteen, while the negative of four squared would be negative sixteen. Losing those brackets is the single commonest way a correct solution appears to fail its check.
Socratic
Throwing away a valid answer sounds wrong.
Discussion prompt
Explain what justifies rejecting the negative solution of a falling object problem. Then say what would be dishonest about rejecting a solution in a pure equation.
Hint: Ask what the variable stands for.
Answer:
The variable stands for a number of seconds since the object was released, and the model only describes what happens after release. A negative value of t refers to a time when the model says nothing, so the solution is outside the domain of the situation rather than wrong in itself — and saying that explicitly is part of a complete answer.
In a pure equation there is no situation to appeal to, so every value satisfying the equation is a solution and discarding one is simply an incomplete answer. The distinction is between a restriction that comes from the problem and a preference that comes from wanting a tidier answer, and only the first is a reason.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Equation | First move | Solutions |
|---|---|---|
| x squared equals 25 | take roots | plus or minus 5 |
| 3x squared minus 48 equals 0 | add 48, then divide by 3 | plus or minus 4 |
| y squared equals -1 | read the sign of d | no real solution |
The middle row is the only one needing real work, and even there the work is Chapter 3 material. The other two rows are answered by looking.
Pattern
For any quadratic equation with no x term, these five moves cover it.
Step three is worth doing before step four rather than after, because it can save the effort of taking a root that does not exist.
OpenStax Elementary Algebra 2e, §10.1 Solve Quadratic Equations Using the Square Root Property §10.1
Check
Both roots.
Check your understanding
Solve x squared equals 121.
Answer: A
Why: Eleven squared and negative eleven squared are both a hundred and twenty-one, so both values solve the equation.
Check
Isolate first.
Check your understanding
Solve 2x squared minus 72 equals 0.
Answer: A
Why: Adding seventy-two and dividing by two gives x squared equals thirty-six, so the solutions are six and negative six.
Check
Read the sign.
Check your understanding
How many real solutions has the equation n squared equals -16?
Answer: A
Why: The square of every real number is positive or nought, so nothing squares to negative sixteen.
Real world
This is the egg-drop contest from the lesson opener. A container is dropped from 32 feet, and the falling object model is negative sixteen t squared plus s.
Discussion prompt
Write the model, find the landing time to the nearest tenth of a second, and then find how long the drop would take from 50 feet. Say which solution you kept each time and why.
Hint: Set the height to nought.
Answer:
\[ 0 = -16t^2 + 32 \;\Longrightarrow\; t^2 = 2 \;\Longrightarrow\; t = \sqrt{2} \approx 1.4 \]
From fifty feet the equation becomes t squared equals 3.125, so the time is about 1.8 seconds. Raising the drop by more than half adds only about four tenths of a second, because the height depends on the square of the time.
In both cases the algebra produced a positive and a negative root and the negative one was discarded, since it would refer to a time before the container was released. Saying that out loud is part of the answer — the equation cannot tell you which of its solutions the world allows.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What is the correct first step in solving 5x squared equals 80?
Correct: Divide both sides by 5.
\[ 5x^2 = 80 \;\to\; x^2 = 16 \;\to\; x = \pm 4 \]
Why: The squared term has to stand completely alone before a root is taken, and the five is a coefficient multiplying it. Dividing gives x squared equals sixteen and then the roots are plus and minus four; checking confirms it, since five times sixteen is eighty. Taking the root first would give the root of five times x, which leaves a radical attached to the variable and makes the rest harder for no reason. Subtracting five treats a multiplier as a term, and dividing by x risks losing a solution and dividing by nought. The order of operations for undoing is the reverse of the order for doing, which is what puts the division ahead of the root.
Explain it
They keep writing only the positive solution.
Discussion prompt
In no more than four sentences, explain why solving by square roots gives two answers. Then give them the check that shows both are genuine.
Hint: Ask what squaring does to a sign.
Answer:
A usable answer: squaring throws away the sign, so both a number and its negative land on the same square. Undoing that cannot tell which one you started from, so both have to be offered — which is what the plus-or-minus sign means.
The check is to substitute each answer into the original equation. Both come out identical once the squaring is done, and seeing the two checks match is the clearest evidence that neither solution is spare.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The plus-or-minus is fixed by writing it as part of the root step itself rather than adding it afterwards. Isolating is fixed by asking whether anything at all is attached to the squared term. The count is fixed by reading the sign of the number on the other side before doing anything. The choosing is fixed by asking what the variable stands for and whether it can be negative. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the standard form of a quadratic equation, circle the leading coefficient, and beneath it write what the form becomes when the middle coefficient is nought. Underneath, write the three cases for the sign of d as three short rules with one example each, and beside them draw a number line showing the two solutions of x squared equals four moving together as d shrinks to nought. In the middle, solve three equations that need rewriting, writing every line and labelling each with the inverse operation used, and check both solutions of one of them in the original equation with the brackets written out. In the lower half, write the falling object model, sketch its graph from a starting height of thirty-two feet, mark the landing point, and solve for the landing time. Beside the graph write one sentence saying which solution you discarded and why. Finally, in the margin, write the reason square roots give two answers while subtraction gives one.
Your checks should show both solutions producing the same line after the squaring step. If they do not, look for a missing pair of brackets around the negative value.
Recap
Five things, and the third can be answered without solving anything.
| If the question says | Your first move is |
|---|---|
| x squared equals a positive number | Take roots, both signs |
| x squared equals a negative number | No real solution; stop |
| There is a coefficient on x squared | Divide it out first |
| An object is dropped from a height | Use h = -16t squared + s |
| Find when it reaches the ground | Set the height to nought |
Lesson 9.3 tidies the radical answers this lesson produced. Expressions like the root of forty-eight can be simplified rather than merely approximated, and doing so makes them far easier to compare and combine.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.2 Solving Quadratic Equations by Finding Square Roots §9.2, pp. 505-510 — everything on these slides traces back here
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