Solve for d 2 square root of d+4=-2

$2 \sqrt{d + 4} = - 2$

Step 1

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

${(2 \sqrt{d + 4})}^{2} = {(- 2)}^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{d + 4}$ as ${(d + 4)}^{\frac{1}{2}}$ .

${(2 {(d + 4)}^{\frac{1}{2}})}^{2} = {(- 2)}^{2}$

Simplify the left side.

Simplify ${(2 {(d + 4)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 {(d + 4)}^{\frac{1}{2}}$ .

$2^{2} {({(d + 4)}^{\frac{1}{2}})}^{2} = {(- 2)}^{2}$

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

$4 {({(d + 4)}^{\frac{1}{2}})}^{2} = {(- 2)}^{2}$

Multiply the exponents in ${({(d + 4)}^{\frac{1}{2}})}^{2}$ .

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

$4 {(d + 4)}^{\frac{1}{2} \cdot 2} = {(- 2)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(d + 4)}^{\frac{1}{2} \cdot 2} = {(- 2)}^{2}$

Rewrite the expression.

$4 {(d + 4)}^{1} = {(- 2)}^{2}$

Simplify.

$4 (d + 4) = {(- 2)}^{2}$

Apply the distributive property.

$4 d + 4 \cdot 4 = {(- 2)}^{2}$

Multiply $4$ by $4$ .

$4 d + 16 = {(- 2)}^{2}$

Simplify the right side.

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

$4 d + 16 = 4$

Step 3

Solve for $d$ .

Move all terms not containing $d$ to the right side of the equation.

Subtract $16$ from both sides of the equation.

$4 d = 4 - 16$

Subtract $16$ from $4$ .

$4 d = - 12$

Divide each term in $4 d = - 12$ by $4$ and simplify.

Divide each term in $4 d = - 12$ by $4$ .

$\frac{4 d}{4} = \frac{- 12}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 d}{4} = \frac{- 12}{4}$

Divide $d$ by $1$ .

$d = \frac{- 12}{4}$

Simplify the right side.

Divide $- 12$ by $4$ .

$d = - 3$

Step 4

Exclude the solutions that do not make $2 \sqrt{d + 4} = - 2$ true.

$No solution$

Solve for d 2 square root of d+4=-2

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