Radical Equations and Extraneous Solutions

7.2: Radical Equations and Extraneous Solutions. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 2 of 83 parts Practice Flashcards

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Isolated radical

Objective: Square a nonnegative radical equation and verify.

Prerequisite skills: square roots, domain conditions, and quadratic equations. For √(x+a)=b, require b≥0 and x+a≥0. Squaring gives x+a=b²; solving yields a candidate that must be substituted into the original equation. Common error: forgetting that a principal root cannot equal a negative number. Restrictions and interpretation: Squaring can create extraneous candidates; all answers must satisfy the original root equation.

Formulas

$√(x+a)=b ⇒ x=b²−a when b≥0$

Worked examples

Try itSolve √(x + 23) = 24.Show answer

The right side 24 is nonnegative. Square both sides: x+23=576, so x=553. Its radicand is 576; the principal root is 24, confirming it. Therefore the requested answer is 553.

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Isolated radical

Objective: Square a nonnegative radical equation and verify.

Formulas

$√(x+a)=b ⇒ x=b²−a when b≥0$

Worked examples

Try itSolve √(x + 23) = 24.Show answer

The right side 24 is nonnegative. Square both sides: x+23=576, so x=553. Its radicand is 576; the principal root is 24, confirming it. Therefore the requested answer is 553.

Extraneous root test

Objective: Discard a squared-equation candidate that fails the original.

Squaring removes sign information. For √(x+a)=x−b, the original requires x≥b in addition to a nonnegative radicand. Solve the quadratic after squaring, then use that sign restriction and direct substitution. Common error: accepting a negative right-hand side because its square matches. Restrictions and interpretation: Squaring can create extraneous candidates; all answers must satisfy the original root equation.

Formulas

$√u = v requires u ≥ 0 and v ≥ 0$

Worked examples

Try itWhat is the real solution of √(x + 577) = x − 23?Show answer

The right side requires x≥23. Write t=x−23; after squaring, t²=t+600, so (t−25)(t+24)=0. Candidates are t=25 and t=-24, giving x=48 or x=-1. The latter violates x≥23. For x=48, the radicand is 625=25² and both sides equal 25; it is valid. Therefore the requested answer is 48.

Radical in a shifted expression

Objective: Use a substitution to preserve the square-root sign.

An equation involving √x can be rewritten with t=√x, where t≥0. Solve in t, reject negative values, then square the surviving t to return to x. This avoids treating √x as an arbitrary signed variable. Common error: squaring a negative t candidate and restoring an invalid solution. Restrictions and interpretation: Squaring can create extraneous candidates; all answers must satisfy the original root equation. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

$t = √x ⇒ t ≥ 0 and x = t²$

Worked examples

Try itWhat is the nonnegative solution of x + √x = 650?Show answer

Let t=√x≥0. Then t²+t−650=0=(t−25)(t+26). Only t=25 is nonnegative, so x=t²=625. Check: 625+25=650. Therefore the requested answer is 625.

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