How do you solve ${(x + 7)}^{\frac{3}{2}} - 20 = 7$ $(x+7)^(3/2)-20=7$ ?

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Answer 1 · 2018-04-22T22:57:15

$x = 2$ $x=2$

${(x + 7)}^{\frac{3}{2}} - 20 = 7$ $(x+7)^(3/2) -20 = 7$

Rearrange equation.

${(x + 7)}^{\frac{3}{2}} = 27$ $(x+7)^(3/2) = 27$

To get rid of the power, first cube root it and then square it.

$({(x + 7)}^{\frac{3}{2} \cdot \frac{1}{3}}) = 27^{\frac{1}{3}}$ $((x+7)^(3/2*1/3)) = 27^(1/3)$

${(x + 7)}^{\frac{3}{6}} = 3 \to {(x + 7)}^{\frac{1}{2}} = 3$ $(x+7)^(3/6) = 3 -> (x+7)^(1/2) = 3$

${(x + 7)}^{\frac{1}{2} \cdot 2} = 3^{2} \to x + 7 = 9$ $(x+7)^(1/2 * 2) = 3^2 -> x+7 = 9$

Rearrange equation to make $x$ $x$ the subject.

$x + 7 = 9 \to x = 9 - 7$ $x+7 = 9 -> x= 9-7$
$x = 2$ $x=2$

How do you solve (x+7)32−20=7(x+7)^(3/2)-20=7?