Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Section 11.1 - Solving Quadratic Equations by Completing the Square - Exercise Set: 28

Answer

$y^2+2y+1= (y+1)^2$

Work Step by Step

We add the square of half of the coefficent of $y$ so that the result is a perfect square trinomial. Co-efficient of $y=2$ Half of 2 is $\frac{1}{2}×2=1$ Square of 1 is $1×1=1$ We add 1 to $y^2+2y$ to make it a perfect square trinomial. Hence it becomes $y^2+2y+1$ Factored form- $y^2+2y+1$ $=y^2+1y+1y+1$ $=y(y+1)+1(y+1)$ (Taking the common factors) $=(y+1)(y+1)$ ($ y+1$ is taken common from both the terms) $=(y+1)^2$
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