## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Zeros: $\dfrac{3+\sqrt{17}}{4}, -\dfrac{3-\sqrt{17}}{4}$ $x$-intercepts: $\dfrac{3+\sqrt{17}}{4}, -\dfrac{3-\sqrt{17}}{4}$
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 $G(x)=0$: $$2x^2-3x-1=0$$ Taking $2$ as a common factor: $$2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)=0$$ $$x^2-\dfrac{3}{2}x-\dfrac{1}{2}=0$$ Rearranging the Equation: $$x^2-\dfrac{3}{2}x=\dfrac{1}{2}$$ The coefficient of $x^2$ is 1 and that of $x$ is $\dfrac{3}{2}$, complete the square by adding $\left(\dfrac{1 \times \dfrac{3}{2}}{2}\right)^2 = \dfrac{9}{16}$ $$\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)=\dfrac{1}{2}+\dfrac{9}{16}$$ $$\left(x-\dfrac{3}{4}\right)^2 = \dfrac{17}{16}$$ $$\left(x-\dfrac{3}{4}\right)=\pm \dfrac{\sqrt{17}}{4}$$ $\therefore x-\dfrac{3}{4} = \dfrac{\sqrt{17}}{4} \hspace{20pt}\text{ or } \hspace{20pt} x-\dfrac{3}{4} = -\dfrac{\sqrt{17}}{4}$ Solve each equation to obtain the zeros of the function (which are the also the function's $x$-intercepts): $x-\dfrac{3}{4} = \dfrac{\sqrt{17}}{4} \to x=\dfrac{3+\sqrt{17}}{4}$ $x-\dfrac{3}{4} = -\dfrac{\sqrt{17}}{4} \to x= \dfrac{3-\sqrt{17}}{4}$