Answer
Zeros: $\dfrac{-\sqrt{2}+2}{2},\dfrac{-\sqrt{2}-2}{2}$
$x$-intercepts: $\dfrac{-\sqrt{2}+2}{2},\dfrac{-\sqrt{2}-2}{2}$
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 $G(x)=0$:
$$x^2+\sqrt{2}x-\dfrac{1}{2}=0$$
Comparing $x^2+\sqrt{2}x-\dfrac{1}{2}=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$
$$\therefore a = 1, b=\sqrt{2}, c =-\dfrac{1}{2}$$
Evaluating the discriminant $b^2-4ac$
$$b^2-4ac = (\sqrt{2})^2-4 \times 1 \times -\dfrac{1}{2} = 4$$
The quadratic formula is given by:
$$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x= \dfrac{-\sqrt{2}\pm \sqrt{4}}{2\times 1}$$
$$x=\dfrac{-\sqrt{2}\pm 2}{2}$$
Thus, the zeros, which are also the $x$-intercepts, are $\dfrac{-\sqrt{2}+2}{2}$ and $\dfrac{-\sqrt{2}-2}{2}$
