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 - Page 165: 19


$x=-4$ 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
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