How do you solve $2 x^{2} - 32 = 0$ $2x^2 - 32 = 0$ ?

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Answer 1 · 2016-04-17T17:11:18

$x = 4, - 4$ $x = 4, -4$

You can factor out a 2 as the GCF.

$2 (x^{2} - 16) = 0$ $2(x^2 - 16) = 0$

$x^{2} - 16$ $x^2 - 16$ is the difference of perfect squares, which are $(x - 4) (x + 4)$ $(x-4)(x+4)$

Therefore, you now have $2 (x - 4) (x + 4)$ $2(x-4)(x+4)$ .

To solve this quadratic means to find the roots of it (the points where the parabola will cross the x-axis).
$x - 4 = 0$ $x - 4 =0$
$x = 4$ $x = 4$

and

$x + 4 = 0$ $x + 4 = 0$
$x = - 4$ $x = -4$

Therefore, x = -4, 4

How do you solve 2x2−32=02x^2 - 32 = 0?