Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.6 Solving Quadratic Equations by Using the Quadratic Formula - 4.6 Exercises - Page 372: 5


$\color{blue}{\left\{-4, 1\right\}}$

Work Step by Step

Recall The solutions of the quadratic equation $ax^2+bx+c=0$ can be found using the quadratic formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ The given quadratic equation has $a=3, b=9, \text{ and } c= -125$. Substitute these values into the quadratic formula to obtain: \begin{align*} x&=\frac{-9\pm\sqrt{9^2-4(3)(-12)}}{2(3)}\\\\ x&=\frac{-9\pm \sqrt{81+144}}{6}\\\\ x&=\frac{-9\pm \sqrt{225}}{6}\\\\ x&=\frac{-9\pm 15}{6}\\\\ \end{align*} Thus, $x_1=\dfrac{-9+15}{6}=\dfrac{6}{6}=1\\\\$ $x_2=\dfrac{-9-15}{6}=\dfrac{-24}{6}=-4$ Therefore, the solution set is $\color{blue}{\left\{-4, 1\right\}}$.
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