1Cademy - Determining if $$ x = 2 $$ and $$ x = -1 $$ are Solutions of $$ \sqrt{x+2} = x $$

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Determining if $x = 2$ and $x = -1$ are Solutions of $\sqrt{x+2} = x$

To determine whether $x = 2$ and $x = -1$ are solutions of the radical equation $\sqrt{x + 2} = x$ , apply the substitution strategy.

Testing $x = 2$ : Substitute $x = 2$ into the left side: $\sqrt{2 + 2} = \sqrt{4} = 2$ . The right side is also $2$ . Since $2 = 2$ is a true statement, $x = 2$ is a valid solution.
Testing $x = -1$ : Substitute $x = -1$ into the left side: $\sqrt{-1 + 2} = \sqrt{1} = 1$ . The right side is $-1$ . Since $1 \neq -1$ , the resulting equation is false. Therefore, $x = -1$ is not a solution; it is an extraneous solution.

The value $x = -1$ fails because it requires the principal square root $\sqrt{1}$ to equal $-1$ , but the radical sign always yields a non-negative result. This example demonstrates why every candidate solution of a radical equation must be verified by substituting it back into the original equation.

Updated 2026-07-02

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OpenStax Elementary Algebra

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