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Algebra 2 Radical functions

Extraneous solutions of radical equations

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Extraneous Solutions of Radical Equations

Texas Algebra II (TEKS) • Standard 2A.4(G) • Radical Functions

Extraneous Solutions of Radical Equations is a topic in Radical Functions in the Texas Essential Knowledge and Skills (Algebra II, §111.40). It is aligned to Standard 2A.4(G), which requires students to identify extraneous solutions of square root equations.

Squaring both sides of a radical equation can introduce extraneous solutions, so every solution must be checked.

Texas Algebra II (TEKS) › Radical Functions › Extraneous Solutions of Radical Equations  —  Standard 2A.4(G)

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Theory

When you square both sides of a radical equation, you can create extraneous solutions — values that satisfy the squared equation but not the original.

Why: squaring loses sign information, since \(a=b\) and \(a=-b\) both give \(a^2=b^2\). A principal square root is never negative.
Extraneous solution from squaring Squaring can add a false solution where the square root and line do not actually meet. x y valid (4,2) extraneous y=√x y=x-2
The false solution is where the curves don't actually meet.
Extraneous solutions Extraneous solutions Extraneous solutions squaring can create false roots a square root is never negative always check in the original
Checking rejects extraneous solutions.

The source of the problem:

\[a=b\ \Rightarrow\ a^2=b^2,\ \text{but not conversely}\]
squaring both sides is not reversible, so it can add solutions
Substitute each solution back into the original equation.

How to handle extraneous solutions

  1. Isolate the radical and square both sides.
  2. Solve the resulting equation.
  3. Substitute each solution into the original.
  4. Keep only those that check.
Example 1 — Reject one solution
Solve \(\sqrt{x}=x-2\).
Solution

Square both sides and solve.

\(x\)\(=\)\((x-2)^2\)
\(x\)\(=\)\(x^2-4x+4\)
\(x^2-5x+4\)\(=\)\(0\)
\(x\)\(=\)\(1,\ 4\)

Check: \(x=1\) fails (\(1\neq-1\)); \(x=4\) works. So \(x=4\).

only x equals 4 works; x equals 1 is extraneous
Example 2 — Another check
Solve \(\sqrt{x+2}=x\).
Solution

Square and solve.

\(x+2\)\(=\)\(x^2\)
\(x^2-x-2\)\(=\)\(0\)
\((x-2)(x+1)\)\(=\)\(0\)
\(x\)\(=\)\(2,\ -1\)

\(x=-1\) fails; \(x=2\) works.

only x equals 2 works
Example 3 — Reject the negative
Solve \(\sqrt{2x+3}=x\).
Solution

Square both sides.

\(2x+3\)\(=\)\(x^2\)
\(x^2-2x-3\)\(=\)\(0\)
\((x-3)(x+1)\)\(=\)\(0\)
\(x\)\(=\)\(3,\ -1\)

\(x=-1\) is extraneous; \(x=3\).

only x equals 3 works
Example 4 — Why they appear
Why does squaring create extraneous solutions?
Solution

Squaring erases sign information: \(a=b\) and \(a=-b\) both give \(a^2=b^2\). A square root is never negative, so some squared solutions don't fit the original.

squaring loses sign information, creating false solutions

Common pitfalls

Always check every solution in the original equation.
A square root can't equal a negative — reject those.
Squaring is not reversible; it can add solutions.

Frequently asked questions

What is an extraneous solution?

A value from the squared equation that fails the original.

Why does squaring cause them?

Squaring loses sign information, so extra values slip in.

How do you find extraneous solutions?

Check each solution in the original equation and discard failures.

Can a square root be negative?

No — the principal square root is always \(\ge 0\).

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