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: 36


The solution set is\[\left\{ -9,\frac{1}{3} \right\}\].

Work Step by Step

Consider the expression\[\left( x+9 \right)\left( 3x-1 \right)=0\]. Then, by the zero product principal either \[\left( x+9 \right)=0\]or\[\left( 3x-1 \right)=0\]. Now \[\left( x+9 \right)=0\]implies that \[x=-9\]and \[\left( 3x-1 \right)=0\] implies that\[x=\frac{1}{3}\]. Next, check the proposed solution by substituting it in the original equation. Check for\[x=-9\]. So consider, \[\begin{align} & \left( x+9 \right)\left( 3x-1 \right)=0 \\ & \left( -9+9 \right)\left( 3\times \left( -9 \right)-1 \right)=0 \\ & 0\left( -28 \right)=0 \\ & 0=0 \end{align}\] Now check for\[x=\frac{1}{3}\]. So consider, \[\begin{align} & \left( x+9 \right)\left( 3x-1 \right)=0 \\ & \left( \frac{1}{3}+9 \right)\left( 3\times \frac{1}{3}-1 \right)=0 \\ & \left( \frac{1}{3}+9 \right)\left( 0 \right)=0 \\ & 0=0 \end{align}\] Hence, the solution set is\[\left\{ -9,\frac{1}{3} \right\}\].
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