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 F - Foundations: A Prelude to Functions - Section F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry - F.2 Assess Your Understanding - Page 16: 2


$\left\{-2, 6\right\}$

Work Step by Step

The quadratic equation $ax^2+bx+c=0$ can be solved using the Quadratic Formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, for the equation: The given equation has $a=1$, $b=-4$, $c=-12$. Using these values and the Quadratic Formula above gives: $x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-12)}}{2(1)}\\ x=\dfrac{4\pm\sqrt{16+48}}{2}\\ x=\dfrac{4\pm\sqrt{64}}{2}\\ x=\dfrac{4\pm8}{2}$ Hence, $x_1=\dfrac{4+8}{2}=6$ $x_2=\dfrac{4-8}{2}=-2$
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