Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 64


$x=\dfrac{1}{4}$ and $x=\dfrac{1}{2}$

Work Step by Step

$x^{2}=\dfrac{3}{4}x-\dfrac{1}{8}$ Take $\dfrac{3}{4}x$ substracting to the left side: $x^{2}-\dfrac{3}{4}x=-\dfrac{1}{8}$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. For this particular problem, $b=-\dfrac{3}{4}$: $x^{2}-\dfrac{3}{4}x+\Big(\dfrac{-3/4}{2}\Big)^{2}=-\dfrac{1}{8}+\Big(\dfrac{-3/4}{2}\Big)^{2}$ $x^{2}-\dfrac{3}{4}x+\dfrac{9}{64}=-\dfrac{1}{8}+\dfrac{9}{64}$ $x^{2}-\dfrac{3}{4}x+\dfrac{9}{64}=\dfrac{1}{64}$ Factor the perfect square trinomial we have on the left side: $\Big(x-\dfrac{3}{8}\Big)^{2}=\dfrac{1}{64}$ Solve for $x$: $x-\dfrac{3}{8}=\pm\sqrt{\dfrac{1}{64}}$ $x-\dfrac{3}{8}=\pm\dfrac{1}{8}$ $x=\dfrac{3}{8}\pm\dfrac{1}{8}$ So, our two solutions are: $x=\dfrac{3}{8}-\dfrac{1}{8}=\dfrac{1}{4}$ and $x=\dfrac{3}{8}+\dfrac{1}{8}=\dfrac{1}{2}$
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