How do you solve $\sqrt[3]{2 x + 4} + 2 = 0$ $root[3]{2x + 4} + 2= 0$ ?

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How do you solve $\sqrt[3]{2 x + 4} + 2 = 0$ $root[3]{2x + 4} + 2= 0$ ?

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Answer 1 · 2016-09-29T19:48:35

$x = - 6$ $x=-6$

Explanation:

$\sqrt[3]{2 x + 4} = - 2$ $root(3)(2x+4) = -2$
${(\sqrt[3]{2 x + 4})}^{3} = {(- 2)}^{3}$ $(root(3)(2x+4))^3 = (-2)^3$
$2 x + 4 = - 8$ $2x+4 = -8$
$2 x = - 12$ $2x = -12$
$x = - 6$ $x=-6$

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How do you solve $\sqrt[3]{2 x + 4} + 2 = 0$ $root[3]{2x + 4} + 2= 0$ ?

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