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: 83


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 $\dfrac{-\sqrt{2}-2}{2}$
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