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: 83

Answer

$x_{1}=0$ and $x_{2}=\dfrac{1}{2}$

Work Step by Step

$(x-3)(3x+4)=(x+2)(x-6)$ $x(3x+4)-3(3x+4)=x(x-6)+2(x-6)$ $3x^2+4x-9x-12=x^2-6x+2x-12$ $3x^2-5x-12=x^2-4x-12$ $3x^2-x^2-5x+4x-12+12=0$ $2x^2-x=0$ $x(2x-1)=0$ Therefore the solutions are $x_{1}=0$ and $x_{2}=\dfrac{1}{2}$
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