Skip to content

Math: Advanced Math, 3 original questions

SAT Nonlinear Equations Practice: Questions and Answers

Solve original SAT-style quadratic, radical and absolute-value equations. Check roots and learn why tempting answer choices fail in three worked questions.

On this page

The short answer

To solve a nonlinear equation, identify its structure before choosing a method. A quadratic may factor, an absolute value may produce two cases, and a radical may require isolating and squaring. Record any restrictions before changing the equation. Operations such as squaring can create candidate solutions that do not satisfy the original statement. Substitute every candidate back, then answer exactly what was requested: an individual root, the number of roots or a combination of them.

Method at a glance

  1. 1Identify the equation type
  2. 2Write restrictions first
  3. 3Solve all relevant cases
  4. 4Verify in the original equation

Try the 3 questionsRead the full method

Step-by-step method

Use these steps in order, then apply them to the questions below.

  1. Step 1: Identify the equation type

    Look for a square, radical, absolute value or denominator. Decide which operation can remove that structure while tracking its conditions.

  2. Step 2: Write restrictions first

    Even square roots are nonnegative, and denominators cannot be zero. These conditions can rule out candidates before or after solving.

  3. Step 3: Solve all relevant cases

    Set a factored quadratic equal to zero, or write both positive and negative cases for a positive absolute-value target.

  4. Step 4: Verify in the original equation

    Check every candidate before calculating the requested sum or choosing a root. A transformed equation can have extra solutions.

Try 3 SAT-style questions

These are original SAT-style questions written by PeakSAT, not official College Board questions. Choose an answer and check it to see why each choice is right or wrong, or open the worked answer.

No sign-in needed. Your answers stay on this page and are not stored.

Question 1 of 3

The equation x² − 9x + 20 = 0 has two real solutions. What is the larger solution?

Choose an answer for question 1
View worked answer for question 1

Correct answer

Choice B: 5

Explanation

Factor the left side as (x − 4)(x − 5), because 4 + 5 = 9 and 4 × 5 = 20. The roots are 4 and 5. The question asks for the larger root, so the answer is 5.

Why each choice is right or wrong

  1. Choice A

    This is a valid root, but it is the smaller one.

  2. Choice B

    Correct answer

    Correct: 5 is the larger of the two roots.

  3. Choice C

    This is the sum of the roots, not an individual solution.

  4. Choice D

    This is their product, not an individual solution.

Question 2 of 3

Which choice gives all real solutions of √(x + 5) = x − 1?

Choose an answer for question 2
View worked answer for question 2

Correct answer

Choice C: 4 only

Explanation

The square root is nonnegative, so x − 1 must be nonnegative and x ≥ 1. Squaring gives x + 5 = (x − 1)², or x² − 3x − 4 = 0. The candidates are 4 and −1. Only 4 satisfies the original equation: √9 = 3.

Why each choice is right or wrong

  1. Choice A

    At x = −1, the two sides are 2 and −2, so they are unequal.

  2. Choice B

    At x = 1, the two sides are √6 and 0, so they are unequal.

  3. Choice C

    Correct answer

    Correct: x = 4 satisfies the equation and the nonnegative-right-side condition.

  4. Choice D

    −1 is an extraneous candidate introduced by squaring; it fails the original equation.

Question 3 of 3

What is the sum of all real solutions of |2x − 3| = 7?

Choose an answer for question 3
View worked answer for question 3

Correct answer

Choice B: 3

Explanation

Write 2x − 3 = 7 or 2x − 3 = −7. The first gives x = 5, and the second gives x = −2. Both satisfy the original equation. Their sum is 5 + (−2) = 3.

Why each choice is right or wrong

  1. Choice A

    This is one root, not the sum of both roots.

  2. Choice B

    Correct answer

    Correct: the two roots sum to 3.

  3. Choice C

    This is the other root, not the requested sum.

  4. Choice D

    This is the absolute-value target, not the sum of the roots.

Common mistakes

  • Stopping after one absolute-value case. A positive target generally requires checking both signs.

  • Accepting every root obtained after squaring. Substitute into the original radical equation to remove extraneous candidates.

  • Reporting a root when the question asks for the larger root or sum. Read the requested quantity after solving.

Keep practicing

Everything on this page is free to use. Practicing this skill in the PeakSAT app requires a subscription; one SAT-style assessment is free.

Subscription

Nonlinear Equations practice in PeakSAT

Sign in to keep practicing this skill with more SAT-style questions and explanations in the PeakSAT app.

Full access requires a subscription.

Continue Nonlinear Equations practiceSee pricing

Free

Take one free SAT-style assessment

See your estimated score and missed questions by skill. Google sign-in, no card required.

Take a free assessment

Common questions

Why do radical equations need a final check?

Squaring preserves a genuine solution but can introduce additional candidates. The original equation determines which candidates are valid.

Can a quadratic have fewer than two real roots?

Yes. It can have two distinct real roots, one repeated real root or no real roots. Its factors or discriminant can distinguish these cases.

Does an absolute-value equation always have two solutions?

No. A negative target has none, and a zero target can have one. For a positive target, solve both signs and check the resulting candidates.

Sources

  1. College Board: Advanced Mathsatsuite.collegeboard.org
  2. College Board: Student Question Bank Math skillssatsuite.collegeboard.org

Sources are linked for reference; they do not endorse PeakSAT. How we prepare our guides