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MathChapter 13: Polynomial and Radical Functions

Notes

Solving Radical Equations Notes

Isolate radicals, raise both sides correctly, solve the resulting equation, and verify every candidate in the original equation.

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A radical equation contains a variable inside a radicand. Raising both sides to a power can remove the radical, but squaring may create candidates that do not satisfy the original equation.

A safe solving workflow

Solve, then verify in the original equationA six-stage workflow isolates a radical, removes it with a power, solves, and checks every candidate before accepting it.
  1. Original equation

    Keep a clean copy for the final check.

  2. Isolate the radical

    Move all nonradical terms away first.

  3. Raise both sides

    Square for a square root or cube for a cube root.

  4. Solve the resulting equation

    Find every algebraic candidate.

  5. Check the original

    Substitute each candidate into the untransformed equation.

  6. Keep or reject

    Retain valid values and label failures extraneous.

Why isolation comes first

  1. Move outside terms

    Turn an equation such as \(4+\sqrt{2x-1}=9\) into \(\sqrt{2x-1}=5\).

  2. Power the entire sides

    Square both complete sides: \((\sqrt{2x-1})^2=5^2\).

  3. Solve and check

    The transformed equation gives candidates; only the original equation decides validity.

Extraneous solutions

Two candidates, one valid solutionThe equation square root of x plus 6 equals x produces candidates 3 and negative 2; original-equation substitution keeps 3 and rejects negative 2.
  1. \(\sqrt{x+6}=x\)

    The right side must be nonnegative.

  2. \(x^2-x-6=0\Rightarrow x=3,-2\)

    Squaring creates two algebraic candidates.

  3. \(\sqrt{3+6}=3\)

    The equality is true, so 3 is valid.

  4. \(\sqrt{-2+6}\ne-2\)

    Two is not negative two, so negative two is extraneous.

Worked example

Check both algebraic candidates

Solve \(\sqrt{x+6}=x\).

  1. Square

    \(x+6=x^2\), so \(x^2-x-6=0\).

  2. Factor

    \((x-3)(x+2)=0\), giving candidates \(3\) and \(-2\).

  3. Check 3

    \(\sqrt{3+6}=3\), so \(3\) is valid.

  4. Check negative 2

    \(\sqrt{-2+6}=2\ne-2\), so \(-2\) is extraneous.

The solution set is \(\{3\}\).

Domain checks and odd roots

Useful checks are not interchangeable

Square-root domain

Require every square-root radicand to be nonnegative. This can eliminate impossible inputs early.

Original substitution

Still substitute every surviving candidate into the original equation; domain membership alone does not prove equality.

Cube-root equations

Cubing is one-to-one over the reals, so it does not create the same sign ambiguity, though arithmetic still needs verification.

Worked example

Cube-root equation

Solve \(\sqrt[3]{2x-5}=-3\).

  1. Cube both sides

    \(2x-5=(-3)^3=-27\).

  2. Solve

    \(2x=-22\), so \(x=-11\).

  3. Verify

    \(\sqrt[3]{2(-11)-5}=\sqrt[3]{-27}=-3\).

\(x=-11\).

Common mistakes and traps

  • Squaring before isolating the radical.
  • Squaring individual terms instead of the entire side.
  • Forgetting the middle term in \((a+b)^2\).
  • Accepting every root of the transformed equation.
  • Checking in the transformed equation rather than the original.
  • Using a domain check as a substitute for equality verification.
Mini check

Verify the candidate

Solve \(\sqrt{3x+1}=4\).

  1. \(3\)
  2. \(5\)
  3. \(15\)
  4. \(17\)
Show answer and explanation

Answer: \(5\)

Squaring gives \(3x+1=16\), so \(x=5\). The original check gives \(\sqrt{16}=4\).

Key takeaways

Key takeaways

What to remember

  • Isolate the radical before applying a power.
  • Squaring may introduce extraneous candidates.
  • Check every candidate in the original equation.
  • Square-root domain restrictions are useful, but they do not replace substitution.
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Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.