How do you solve $4 x^{2} + 32 = 0$ $4x^2+32=0$ ?

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How do you solve $4 x^{2} + 32 = 0$ $4x^2+32=0$ ?

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Answer 1 · 2016-10-05T02:45:28

First simplification is by dividing by $4$ $4$

Explanation:

$x^{2} + 8 = 0 \to x^{2} = - 8 \to$ $x^2+8=0->x^2=-8->$
$x = \pm \sqrt{- 8} = \pm \sqrt{(- 1) \cdot 8} \to$ $x=+-sqrt(-8)=+-sqrt((-1)*8)->$
$x = \pm \sqrt{- 1} \cdot \sqrt{8} = \pm i \sqrt{8}$ $x=+-sqrt(-1)*sqrt8=+-isqrt8$

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How do you solve $4 x^{2} + 32 = 0$ $4x^2+32=0$ ?

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How do you solve $4 x^{2} + 32 = 0$ $4x^2+32=0$ ?