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: 86

Answer

Zeros: $\dfrac{-1+\sqrt{5}}{2},\dfrac{-1-\sqrt{5}}{2}$ $x$-intercepts: $\dfrac{-1+\sqrt{5}}{2},\dfrac{-1-\sqrt{5}}{2}$

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 $g(x)=0$: $$x^2+x-1=0$$ Comparing $x^2+x-1=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$ $$\therefore a =1, b=1, c =-1$$ Evaluating the discriminant $b^2-4ac$ $$b^2-4ac = (-1)^2-4 \times 1\times -1 = 5$$ The quadratic formula is given by: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{-1\pm \sqrt{5}}{2\times 1}$$ $$x=\dfrac{-1\pm \sqrt{5}}{2}$$ Thus, the zeros, which are also the $x$-intercepts, are $\dfrac{-1+\sqrt{5}}{2}$ and $\dfrac{-1-\sqrt{5}}{2}$.
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