Solve: \sqrt{x+12} = x. Identify any extraneous solutions.

x = 4 only; x = -3 extraneous x + 12 = x^2 \Rightarrow x^2 - x - 12 = 0 \Rightarrow (x-4)(x+3) = 0. Check x = -3: \sqrt{9} = 3 \neq -3, extraneous. Valid: x = 4.

Nonlinear Equations in One Variable and Systems of Equations

SAT Math· difficulty 4/5

\sqrt{x+12} = x

A
$x = 4$
check_circle
B
$No solution$
C
$x = -3$
D
$x = 4$

Explanation

x + 12 = x^2 \Rightarrow x^2 - x - 12 = 0 \Rightarrow (x-4)(x+3) = 0

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