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 399: 33


\[\left\{ -3,8 \right\}\].

Work Step by Step

Consider the expression\[\left( x-8 \right)\left( x+3 \right)=0\]. Then, by the zero product principal either \[\left( x-8 \right)=0\]or\[\left( x+3 \right)=0\]. Now \[\left( x-8 \right)=0\]implies that \[x=8\]and \[\left( x+3 \right)=0\] implies that\[x=-3\]. Next check the proposed solution by substituting it in the original equation. Check for\[x=8\]. So consider, \[\begin{align} & \left( x-8 \right)\left( x+3 \right)=0 \\ & \left( 8-8 \right)\left( 8+3 \right)=0 \\ & 0\left( 11 \right)=0 \\ & 0=0 \end{align}\] Now check for\[x=-3\]. So consider, \[\begin{align} & \left( x-8 \right)\left( x+3 \right)=0 \\ & \left( -3-8 \right)\left( -3+3 \right)=0 \\ & \left( -11 \right)0=0 \\ & 0=0 \end{align}\] Hence, the solution set is\[\left\{ -3,8 \right\}\].
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