Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.1 - Solving Quadratic Equations by Completing the Square - Exercise Set - Page 765: 58



Work Step by Step

$4x^{2}-2x+5=0$ Take the $5$ to the right side of the equation: $4x^{2}-2x=-5$ Take out common factor $4$ from the left side of the equation: $4\Big(x^{2}-\dfrac{1}{2}x\Big)=-5$ Take the $4$ to divide the right side of the equation: $x^{2}-\dfrac{1}{2}x=-\dfrac{5}{4}$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-\dfrac{1}{2}$ $x^{2}-\dfrac{1}{2}x+\Big(\dfrac{-\frac{1}{2}}{2}\Big)^{2}=-\dfrac{5}{4}+\Big(\dfrac{-\frac{1}{2}}{2}\Big)^{2}$ $x^{2}-\dfrac{1}{2}x+\dfrac{1}{16}=-\dfrac{5}{4}+\dfrac{1}{16}$ $x^{2}-\dfrac{1}{2}x+\dfrac{1}{16}=-\dfrac{19}{16}$ Factor the left side of the equation, which is a perfect square trinomial: $\Big(x-\dfrac{1}{4}\Big)^{2}=-\dfrac{19}{16}$ Take the square root of both sides of the equation: $\sqrt{\Big(x-\dfrac{1}{4}\Big)^{2}}=\sqrt{-\dfrac{19}{16}}$ $x-\dfrac{1}{4}=\pm\dfrac{\sqrt{19}}{4}i$ Solve for $x$: $x=\dfrac{1\pm\sqrt{19}i}{4}$
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.