Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Section 11.3 - Integrated Review - Summary on Solving Quadratic Equations: 16

Answer

$x_{1}= \dfrac{1 + \sqrt{13}}{4}$ and $x_{2} = \dfrac{1 - \sqrt{13}}{4}$

Work Step by Step

Given $4x^2-2x-3=0$ $a= 4, \ b=-2, \ c=-3$ Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have: $\dfrac{-(-2) \pm \sqrt{(-2)^2-4 \times 4\times (-3)}}{2 \times 4} = \dfrac{2 \pm \sqrt{4+48}}{8} = \dfrac{2 \pm \sqrt{52}}{8} = \dfrac{2 \pm 2\sqrt{13}}{8} = \dfrac{1 \pm \sqrt{13}}{4}$ Therefore the solutions are $x_{1}= \dfrac{1 + \sqrt{13}}{4}$ and $x_{2} = \dfrac{1 - \sqrt{13}}{4}$
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.