Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 63


The solutions are $x=-\dfrac{3}{8}\pm\dfrac{\sqrt{41}}{8}$

Work Step by Step

$(4x-1)(x+2)=4x$ Evaluate the product on the left side: $4x^{2}+8x-x-2=4x$ Take $4x$ to the left side and simplify: $4x^{2}+8x-x-4x-2=0$ $4x^{2}+3x-2=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=4$, $b=3$ and $c=−2$ Substitute the known values into the formula and evaluate: $x=\dfrac{-3\pm\sqrt{3^{2}-4(4)(-2)}}{2(4)}=\dfrac{-3\pm\sqrt{9+32}}{8}=...$ $...=\dfrac{-3\pm\sqrt{41}}{8}=-\dfrac{3}{8}\pm\dfrac{\sqrt{41}}{8}$ The solutions are $x=-\dfrac{3}{8}\pm\dfrac{\sqrt{41}}{8}$
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