Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 99

Answer

$x=\frac{-11\pm\sqrt {33}}{4}$

Work Step by Step

$ (2x + 3)(x + 4) = 1$ $2x^2+11x+12=1$ $2x^2+11x+11=0$ $x=\frac{-11\pm\sqrt {{11}^2-4*2*11}}{2*2}=\frac{-11\pm\sqrt {121-88}}{4}=\frac{-11\pm\sqrt {33}}{4}$
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