Solve for h 2 square root of h=-8

$2 \sqrt{h} = - 8$

To remove the radical on the left side of the equation, square both sides of the equation.

${(2 \sqrt{h})}^{2} = {(- 8)}^{2}$

Simplify each side of the equation.

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Apply the product rule to $2 h^{\frac{1}{2}}$ .

$2^{2} {(h^{\frac{1}{2}})}^{2} = {(- 8)}^{2}$

Raise $2$ to the power of $2$ .

$4 {(h^{\frac{1}{2}})}^{2} = {(- 8)}^{2}$

Multiply the exponents in ${(h^{\frac{1}{2}})}^{2}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 h^{\frac{1}{2} \cdot 2} = {(- 8)}^{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$4 h^{\frac{1}{2} \cdot 2} = {(- 8)}^{2}$

Rewrite the expression.

$4 h^{1} = {(- 8)}^{2}$

Simplify.

$4 h = {(- 8)}^{2}$

Raise $- 8$ to the power of $2$ .

$4 h = 64$

Divide each term by $4$ and simplify.

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Divide each term in $4 h = 64$ by $4$ .

$\frac{4 h}{4} = \frac{64}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 h}{4} = \frac{64}{4}$

Divide $h$ by $1$ .

$h = \frac{64}{4}$

Divide $64$ by $4$ .

$h = 16$

Exclude the solutions that do not make $2 \sqrt{h} = - 8$ true.

No solution

Solve for h 2 square root of h=-8

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