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

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 2 - Linear and Quadratic Functions - Section 2.3 Quadratic Functions and Their Zeros - 2.3 Assess Your Understanding - Page 146: 36


Zeros: $\dfrac{3+\sqrt{17}}{4}, -\dfrac{3-\sqrt{17}}{4}$ $x$-intercepts: $\dfrac{3+\sqrt{17}}{4}, -\dfrac{3-\sqrt{17}}{4}$

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 $G(x)=0$: $$2x^2-3x-1=0$$ 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} $ $x-\dfrac{3}{4} = -\dfrac{\sqrt{17}}{4} \to x= \dfrac{3-\sqrt{17}}{4}$
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