Solve for d 3- square root of d-4=0

$3 - \sqrt{d - 4} = 0$

Step 1

Subtract $3$ from both sides of the equation.

$- \sqrt{d - 4} = - 3$

Step 2

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

${(- \sqrt{d - 4})}^{2} = {(- 3)}^{2}$

Step 3

Simplify each side of the equation.

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

${(- {(d - 4)}^{\frac{1}{2}})}^{2} = {(- 3)}^{2}$

Simplify the left side.

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

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

${(- 1)}^{2} {({(d - 4)}^{\frac{1}{2}})}^{2} = {(- 3)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(d - 4)}^{\frac{1}{2}})}^{2} = {(- 3)}^{2}$

Multiply ${({(d - 4)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(d - 4)}^{\frac{1}{2}})}^{2} = {(- 3)}^{2}$

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

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

${(d - 4)}^{\frac{1}{2} \cdot 2} = {(- 3)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(d - 4)}^{\frac{1}{2} \cdot 2} = {(- 3)}^{2}$

Rewrite the expression.

${(d - 4)}^{1} = {(- 3)}^{2}$

Simplify.

$d - 4 = {(- 3)}^{2}$

Simplify the right side.

Raise $- 3$ to the power of $2$ .

$d - 4 = 9$

Step 4

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

Add $4$ to both sides of the equation.

$d = 9 + 4$

Add $9$ and $4$ .

$d = 13$