Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - 6.5 Quadratic Equations - Exercise Set 6.5 - Page 400: 61

Answer

the solution set is\[\left\{ -3,6 \right\}\].

Work Step by Step

The given equation can be written as \[{{x}^{2}}-3x-18=0\]. Compare the given equation with the equation,\[a{{x}^{2}}+bx+c=0\]. Here,\[a=1,\,\,b=-3,\text{ and }c=-18\] Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] That is, \[\begin{align} & x=\frac{3\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times \left( -18 \right)}}{2\times 1} \\ & =\frac{3\pm \sqrt{9+72}}{2} \\ & =\frac{3\pm \sqrt{81}}{2} \\ & =\frac{3\pm 9}{2} \end{align}\] Further simplify it to get, \[\begin{align} & x=\frac{3\pm 9}{2} \\ & =\frac{3-9}{2},\frac{3+9}{2} \\ & =-\frac{6}{2},\frac{12}{2} \\ & =-3,6 \end{align}\] Hence, the solution set is\[\left\{ -3,6 \right\}\].
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