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: 40


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: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{-6 \pm \sqrt{32}}{2\times 1}$$ $$x=\dfrac{-6 \pm 4\sqrt{2}}{2}$$ $$x= -3\pm 2\sqrt{2}$$ $\therefore x =-3+2\sqrt{2} \hspace{20pt} \text{or} \hspace{20pt} x=-3- 2\sqrt{2}$
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