Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - Chapter Review - Page 320: 23



Work Step by Step

$3x^2-24x-48=0\\3(x^2-8x-16)=0\\x^2-8x-16=0$ The quadratic formula says that for $ax^2+bx+x$: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Hence here $x=\frac{8\pm\sqrt{(-8)^2-4(1)(-16)}}{2(1)}=\frac{8\pm\sqrt{64+64}}{2}=\frac{8\pm\sqrt{128}}{2}=\frac{8\pm8\sqrt{2}}{2}=4\pm4\sqrt2$
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