How do you solve the equation $- 4 (x^{2} - 8) = 16$ $-4(x^2-8)=16$ ?

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Answer 1 · 2017-05-21T23:36:22

$x = + 2 or x = - 2$ $x = +2 or x=-2$

There is no $x$ $x$ =term, so we can solve this by finding the square roots.

$- 4 (x^{2} - 8) = 16 \leftarrow \div - 4$ $-4(x^2-8)=16" "larr div -4$

$x^{2} - 8 = - 4 \leftarrow$ $x^2 -8= -4" "larr$ add 8 to both sides

$x^{2} = - 4 + 8$ $x^2 = -4+8$

$x^{2} = 4$ $x^2 = 4$

$x = \pm \sqrt{4}$ $x = +-sqrt4$

$x = \pm 2$ $x =+-2$

How do you solve the equation −4(x2−8)=16-4(x^2-8)=16?