Answer
Zeros: $2\sqrt{3}-2,-2\sqrt{3}-2$
$x$-intercepts: $2\sqrt{3}-2,-2\sqrt{3}-2$
Work Step by Step
To find the zeros of a function $g$, solve the equation $g(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $f(x)=0$:
$$x^2+4x-8=0$$
Rearranging the Equation:
$$x^2+4x=8$$
The coefficient of $x^2$ is 1 and that of x is 4, complete the square by adding $\left(\dfrac{1 \times 4}{2}\right)^2 = 4$
$$(x^2+4x+4)=8+4$$
$$(x+2)^2 = 12$$
$$(x+2)=\pm \sqrt{12}$$
$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}$
$\therefore x+2 = 2\sqrt{3} \hspace{20pt}\text{ or } \hspace{20pt} x+2 = -2 \sqrt{3} $
Solve each equation to obtain the zeros of the given function:
$x+2 = 2 \sqrt{3} \to x=2\sqrt{3}-2$
$x+2 = -2 \sqrt{3} \to x= -2 \sqrt{3}-2$
Thus, the $x$-intercepts are: $-2\sqrt3-2$ and $2\sqrt3-2$
