Why is -4 an invalid root for the equation? -2 is also a root of +4 right? Question is taken from the GRE Quant Problem Solving on Gregmat website.

(x+2)^2= x+8

x^2+4+4x=x+8

x^2-4+3x=0

x=1,-4

Now, verfying the solution

First : x =1

(x+2)= \sqrt{x+8}

(1+2)= \sqrt{1+8}

3=3

LHS = RHS

Second x = -4

(x+2)= \sqrt{x+8}

(-4+2)= \sqrt{-4+8}

-2\neq2

LHS \neq RHS

Thus, only x = 1 is our solution

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