Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 391: 61



Work Step by Step

As per given question: $3x-5=(x-1)^2$ This can be written as: $3x-5=x^2-2x+1$ or, $x^2-2x+1-3x+5=0$ or, $x^2-5x+6=0$ Need to use factorization to solve the quadratic equation . $x^2-2x-3x-6=0$ or, $(x-2)(x-3)=0$ This implies $x-2=0$ and $x-3=0$ Hence, $x=2,3$
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