Solve the equation using square roots

2x^2-128=0
A. -sqrt8, sqrt 8
B. -64 or 64
C. -8 or 8
D. No real number solutions

I am stuck between A and C and i also have another problem similer to this one and its the same thing i am confused as do i leave the sqaure root on or no.

1 answer

2x^2 - 128 = 0
x^2 = 64
x = ±√64 = ±8
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