Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 6


quadratic x+1 $W^{2}-5W+6=0$

Work Step by Step

$(x+1)^2-5(x+1)+6=0$ Remember Linear takes form of $ax+b=0$ quadratic takes form of $ax^{2}+bx+c=0$ so this is quadratic type Set W=(x+1) [substitute (x+1) with W] $W^2-5W+6=0$
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