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: 56



Work Step by Step

Given: $f(c)=5c^2-11c+6$ Plug $f(c)=4$ then we have $5c^2-11c+6=4$ This can be written as: $5c^2-11c+2=0$ Need to use factorization to solve the quadratic equation . $5c^2-10c-c+2=0$ This implies $(c-2)(5c-1)=0$ Now, $c-2=0 $ and $5c-1=0$ Hence, $c=2,\dfrac{1}{5}$
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