How do you solve the equation $4 {(x + 5)}^{2} = - 64$ $4(x+5)^2=-64$ ?

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How do you solve the equation $4 {(x + 5)}^{2} = - 64$ $4(x+5)^2=-64$ ?

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Answer 1 · 2017-06-20T07:03:25

$x = - 5 \pm 4 i$ $x=-5+-4i$

Explanation:

$Isolate {(x + 5)}^{2} by dividing both sides by 4$ $"Isolate " (x+5)^2" by dividing both sides by 4"$

$\Rightarrow {(x + 5)}^{2} = - 16$ $rArr(x+5)^2=-16$

${(x + 5)}^{2} \geq 0 hence there are no real roots$ $(x+5)^2>=0" hence there are no real roots"$

$x + 5 = \pm \sqrt{- 16}$ $x+5=+-sqrt(-16)$

$x + 5 = \pm 4 i$ $color(white)(x+5)=+-4i$

$\Rightarrow x = - 5 \pm 4 i$ $rArrx=-5+-4i$

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How do you solve the equation $4 {(x + 5)}^{2} = - 64$ $4(x+5)^2=-64$ ?

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How do you solve the equation $4 {(x + 5)}^{2} = - 64$ $4(x+5)^2=-64$ ?