Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-5 Completing the Square - Practice and Problem-Solving Exercises - Page 564: 23



Work Step by Step

$t^2+t-28$ or, $t^2+t=28$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=1$ Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$ Thus, $c=\dfrac{b^2}{4a}=\dfrac{(1)^2}{4}=\dfrac{1}{4}$ To complete the square, add $\dfrac{1}{4}$ on both sides. $t^2+t+\dfrac{1}{4}=28+\dfrac{1}{4}$ $\implies (t+\dfrac{1}{2})^2=\dfrac{113}{4}$ $\implies (t+\dfrac{1}{2})=5.32$ and $\implies (t+\dfrac{1}{2})=-5.32$ or, $t=4.82,-5.82$
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