Answer
Zeros: $-3+2\sqrt{2},-3-2\sqrt{2}$
$x$-intercepts: $-3+2\sqrt{2},-3-2\sqrt{2}$
Work Step by Step
To find the zeros of a function $f$, solve the equation $f(x)=0$
The zeros of the function are also the $x-$intercepts.
Let $g(x)=0$:
$$x^2+6x+1=0$$
Comparing $x^2+6x+1$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 1, b=6 , c =1$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (6)^2-4 \times 1 \times 1 = 32$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{-6 \pm \sqrt{32}}{2\times 1}$$
$$x=\dfrac{-6 \pm 4\sqrt{2}}{2}$$
$$x= -3\pm 2\sqrt{2}$$
$\therefore x =-3+2\sqrt{2} \hspace{20pt} \text{or} \hspace{20pt} x=-3- 2\sqrt{2}$
