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: 40


The solution set is\[\left\{ -7,6 \right\}\].

Work Step by Step

Step 1:Shift all nonzero terms to left side and obtain zero on the other side. Since all the nonzero terms of the given expression are already on the left side, so no need to do step one. Step 2:Find the factor of the above equation: Consider the equation\[{{x}^{2}}+x-42=0\]. Factorize it as follows: \[\begin{align} & {{x}^{2}}+x-42=0 \\ & {{x}^{2}}+7x-6x-42=0 \\ & x\left( x+7 \right)-6\left( x+7 \right)=0 \\ & \left( x+7 \right)\left( x-6 \right)=0 \end{align}\] Steps 3 and 4: Set each factor equal to zero and solve the resulting equation: From step 2, \[\left( x+7 \right)\left( x-6 \right)=0\]. By the zero product principal, either \[\left( x+7 \right)=0\]or\[\left( x-6 \right)=0\]. Now \[\left( x+7 \right)=0\] implies that\[x=-7\] and \[\left( x-6 \right)=0\]implies that\[x=6\]. Step5: Check the solution in the original equation. Check for\[x=-7\]. So consider, \[\begin{align} & {{x}^{2}}+x-42=0 \\ & {{\left( -7 \right)}^{2}}+\left( -7 \right)-42=0 \\ & 49-7-42=0 \\ & 0=0 \end{align}\] And, check for\[x=6\]. \[\begin{align} & {{x}^{2}}+x-42=0 \\ & {{\left( 6 \right)}^{2}}+\left( 6 \right)-42=0 \\ & 36+6-42=0 \\ & 0=0 \end{align}\] Hence, the solution set is\[\left\{ -7,6 \right\}\].
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