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: 34

Answer

$p^2-7p+\frac{49}{4}=(p-\frac{7}{2})^2$

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 $p=-7$ Half of -7 is $\frac{1}{2}×-7=\frac{-7}{2}$ Square of $\frac{-7}{2}$ is $\frac{-7}{2}×\frac{-7}{2}=\frac{49}{4}$ We add $\frac{49}{4}$ to $p^2-7p$ to make it a perfect square trinomial. Hence it becomes $p^2-7p+\frac{49}{4}$ Factored form- $p^2-7p+\frac{49}{4}$ $= p^2-\frac{7}{2}p-\frac{7}{2}p +\frac{49}{4} $ $=p(p-\frac{7}{2})-\frac{7}{2} (p-\frac{7}{2})$ (Taking the common factors) $=(p-\frac{7}{2}) (p-\frac{7}{2})$ ($(p-\frac{7}{2})$ is taken common from both the terms) $=(p-\frac{7}{2})^2$
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