How do you solve #2x^2 - 32 = 0#?

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Answer 1 · 2016-04-17T17:11:18

#x = 4, -4#

You can factor out a 2 as the GCF.

#2(x^2 - 16) = 0#

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

Therefore, you now have #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#

and

#x + 4 = 0#
#x = -4#

Therefore, x = -4, 4