College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 160: 61


The solutions are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{2}}{2}$

Work Step by Step

$4x^{2}-4x-1=0$ Take out common factor $4$ from the left side: $4\Big(x^{2}-x-\dfrac{1}{4}\Big)=0$ Take the $4$ to divide the right side: $x^{2}-x-\dfrac{1}{4}=\dfrac{0}{4}$ $x^{2}-x-\dfrac{1}{4}=0$ Take $\dfrac{1}{4}$ to the right side: $x^{2}-x=\dfrac{1}{4}$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-1$ $x^{2}-x+\Big(\dfrac{-1}{2}\Big)^{2}=\dfrac{1}{4}+\Big(\dfrac{-1}{2}\Big)^{2}$ $x^{2}-x+\dfrac{1}{4}=\dfrac{1}{4}+\dfrac{1}{4}$ $x^{2}-x+\dfrac{1}{4}=\dfrac{1}{2}$ Factor the left side of the equation, which is a perfect square trinomial: $\Big(x-\dfrac{1}{2}\Big)^{2}=\dfrac{1}{2}$ Take the square root of both sides: $\sqrt{\Big(x-\dfrac{1}{2}\Big)^{2}}=\sqrt{\dfrac{1}{2}}$ $x-\dfrac{1}{2}=\pm\dfrac{1}{\sqrt{2}}$ $x-\dfrac{1}{2}=\pm\dfrac{\sqrt{2}}{2}$ Solve for $x$: $x=\dfrac{1}{2}\pm\dfrac{\sqrt{2}}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.