Find the Roots (Zeros) f(x) = square root of x+4-3

$f (x) = \sqrt{x + 4} - 3$

Set $\sqrt{x + 4} - 3$ equal to $0$ .

$\sqrt{x + 4} - 3 = 0$

Solve for $x$ .

Add $3$ to both sides of the equation.

$\sqrt{x + 4} = 3$

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

${\sqrt{x + 4}}^{2} = 3^{2}$

Simplify each side of the equation.

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

${({(x + 4)}^{\frac{1}{2}})}^{2} = 3^{2}$

Simplify the left side.

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

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

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

${(x + 4)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(x + 4)}^{\frac{1}{2} \cdot 2} = 3^{2}$

Rewrite the expression.

${(x + 4)}^{1} = 3^{2}$

Simplify.

$x + 4 = 3^{2}$

Simplify the right side.

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

$x + 4 = 9$

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

Subtract $4$ from both sides of the equation.

$x = 9 - 4$

Subtract $4$ from $9$ .

$x = 5$