College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Mid-Chapter Check Point: 19

Answer

$x=-4$ and $x=\dfrac{1}{2}$ satisfy the given conditions

Work Step by Step

$y_{1}=2x+3$ $,$ $y_{2}=x+2$ and $y_{1}y_{2}=10$ If $y_{1}y_{2}$ is equal to $10$, then $(2x+3)(x+2)$ is also equal to $10$ $(2x+3)(x+2)=10$ Evaluate the product on the left side: $2x^{2}+4x+3x+6=10$ Take the $10$ to the left side and simplify: $2x^{2}+4x+3x+6-10=0$ $2x^{2}+7x-4=0$ Solve by factoring: $(x+4)(2x-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+4=0$ $x=-4$ $2x-1=0$ $2x=1$ $x=\dfrac{1}{2}$ $x=-4$ and $x=\dfrac{1}{2}$ satisfy the given conditions
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.