Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set: 85


$x_{1}=0$ and $x_{2}=-15$

Work Step by Step

$(2x-3)(x+8)=(x-6)(x+4)$ $2x(x+8)-3(x+8)=x(x+4)-6(x+4)$ $2x^2+16x-3x-24=x^2+4x-6x-24$ $2x^2+13x-24=x^2-2x-24$ $2x^2-x^2+13x+2x-24+24=0$ $x^2+15x=0$ $x(x+15)=0$ Therefore the solutions are $x_{1}=0$ and $x_{2}=-15$
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