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: 31

Answer

$p^2+9p+\frac{81}{4}=(p+\frac{9}{2})^2$

Work Step by Step

We add the square of half of the coefficent of $p$ so that the result is a perfect square trinomial. Co-efficient of $p=9$ Half of 9 is $\frac{1}{2}×9=\frac{9}{2}$ Square of $\frac{9}{2}$ is $\frac{9}{2}×\frac{9}{2}=\frac{81}{4}$ We add $\frac{81}{4}$ to $p^2+9p$ to make it a perfect square trinomial. Hence it becomes $p^2+9p+\frac{81}{4}$ Factored form- $p^2+9p+\frac{81}{4}$ $= p^2+\frac{9}{2}p+\frac{9}{2}p +\frac{81}{4} $ $=p(p+\frac{9}{2})+\frac{9}{2} (p+\frac{9}{2})$ (Taking the common factors) $=(p+\frac{9}{2}) (p+\frac{9}{2})$ ($(p+\frac{9}{2})$ is taken common from both the terms) $=(p+\frac{9}{2})^2$
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