Thinking Mathematically (6th Edition)

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

Chapter 6 - Algebra: Equations and Inequalities - Chapter Summary, Review, and Test - Review Exercises - Page 405: 49

Answer

The solution set is\[\left\{ -8,\frac{1}{2} \right\}\].

Work Step by Step

Step1: 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. Step2: Find the factor of the above equation: Consider the equation,\[2{{x}^{2}}+15x-8=0\] Factorize it as follows: \[\begin{align} & 2{{x}^{2}}+15x-8=0 \\ & 2{{x}^{2}}+16x-x-8=0 \\ & 2x\left( x+8 \right)-1\left( x+8 \right)=0 \\ & \left( 2x-1 \right)\left( x+8 \right)=0 \end{align}\] Step3 and 4: Set each factor equal to zero and solve the resulting equation: From Step 2, \[\left( 2x-1 \right)\left( x+8 \right)=0\] By the zero product principal either \[\left( 2x-1 \right)=0\]or\[\left( x+8 \right)=0\]. Now \[\left( 2x-1 \right)=0\] implies that\[x=\frac{1}{2}\] and, \[\left( x+8 \right)=0\]implies that\[x=-8\]. Step5: Check the solution in the original equation: Check for\[x=-8\]. So consider, \[\begin{align} & 2{{x}^{2}}+15x-8=0 \\ & 2{{\left( -8 \right)}^{2}}+15\left( -8 \right)-8=0 \\ & 128-120-8=0 \\ & 0=0 \end{align}\] and, check for\[x=\frac{1}{2}\]. \[\begin{align} & 2{{x}^{2}}+15x-8=0 \\ & 2{{\left( \frac{1}{2} \right)}^{2}}+15\left( \frac{1}{2} \right)-8=0 \\ & \frac{1}{2}+\frac{15}{2}-8=0 \\ & 0=0 \end{align}\]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.