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: 8

Answer

$121$

Work Step by Step

$z^2+22z+c$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=22$ It will have discriminant, $d$ with formula : $d=b^2-4ac$ Since, $d=0$ for repeated root of the expression. Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$ To complete the square, plug in $a=1, b=22$. Thus, $c=\dfrac{b^2}{4a}=\dfrac{(22)^2}{4}=121$
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