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: 4


$\color{blue}{\left\{-3, 15\right\}}$

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 $a=1, b=-12, \text{ and } c= -45$. Substitute these values into the quadratic formula to obtain: \begin{align*} x&=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(-45)}}{2(1)}\\\\ x&=\frac{12\pm \sqrt{144+180}}{2}\\\\ x&=\frac{12\pm \sqrt{324}}{2}\\\\ x&=\frac{12\pm 18}{2}\\\\ \end{align*} Thus, $x_1=\dfrac{12+18}{2}=\dfrac{30}{2}=15\\\\$ $x_2=\dfrac{12-18}{2}=\dfrac{-6}{2}=-3$ Therefore, the solution set is $\color{blue}{\left\{-3, 15\right\}}$.
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