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Algebra I to AP Precalculus · Module 9 of 14

Radicals & Absolute Value

Simplifying square roots, solving radical and absolute value equations, and absolute value inequalities.

In this module
  1. Simplifying Radical Expressions
  2. Solving Radical Equations
  3. Absolute Value Equations
  4. Absolute Value Inequalities
  5. Check yourself

Simplifying Radical Expressions

  • √a · √b = √(ab) for a, b ≥ 0 — pull out perfect-square factors to simplify: rewrite the number under the root as (perfect square) × (leftover), then take the square root of the perfect-square part.
  • A radical is fully simplified when no perfect-square factor remains under the root.
Worked example

Simplify √72.

Solution. 72 = 36 × 2, and 36 is a perfect square. √72 = √36 · √2 = 6√2.

Solving Radical Equations

  • Isolate the radical on one side, then square both sides to eliminate it, then solve the resulting equation normally.
  • Squaring both sides can introduce extraneous solutions (values that satisfy the squared equation but not the original) — always substitute your answer back into the original equation to check.
Worked example

Solve √(x + 3) = 5.

Solution. Square both sides: x + 3 = 25, so x = 22. Check: √(22 + 3) = √25 = 5 ✓. Solution: x = 22.

Common mistake: Squaring both sides of a false equation can produce a true one. For example, squaring √(3x + 1) = −2 gives 3x + 1 = 4, so x = 1 — but √(3(1) + 1) = √4 = 2, not −2. Since a principal square root can never equal a negative number, x = 1 is extraneous and the equation actually has no solution.

Absolute Value Equations

  • |x| is the distance from 0 on the number line, so it's always ≥ 0. Solving |expression| = k for k > 0 means expression = k OR expression = −k.
  • If k < 0, the equation |expression| = k has no solution, since an absolute value can never be negative.
Worked example

Solve |2x − 3| = 7.

Solution. Case 1: 2x − 3 = 7 → 2x = 10 → x = 5. Case 2: 2x − 3 = −7 → 2x = −4 → x = −2. Solutions: x = 5 or x = −2.

Absolute Value Inequalities

  • |expression| < k (for k > 0) becomes the compound "and" inequality −k < expression < k.
  • |expression| > k (for k > 0) becomes the compound "or" inequality expression < −k OR expression > k.
Worked example

Solve |x − 4| < 6.

Solution. Rewrite as a compound inequality: −6 < x − 4 < 6. Add 4 to all three parts: −2 < x < 10.

Check yourself

2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.

1. Simplify √48.

2. Solve √(x − 2) = 4.

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