Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.5 - Graphs of Quadratic Equations - Exercise Set: 9


$\text{$x-$intercepts: } \{ -1-\sqrt{5},-1+\sqrt{5} \} $

Work Step by Step

Setting $y=0$ and solving for $x$ using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), then the $x-$intercept/s of the given equation, $ y=x^2+2x-4 ,$ is/are \begin{array}{l}\require{cancel} x^2+2x-4=0 \\\\ x=\dfrac{-2\pm\sqrt{2^2-4(1)(-4)}}{2(1)} \\\\ x=\dfrac{-2\pm\sqrt{4+16}}{2} \\\\ x=\dfrac{-2\pm\sqrt{20}}{2} \\\\ x=\dfrac{-2\pm\sqrt{4\cdot5}}{2} \\\\ x=\dfrac{-2\pm\sqrt{(2)^2\cdot5}}{2} \\\\ x=\dfrac{-2\pm2\sqrt{5}}{2} \\\\ x=\dfrac{2(-1\pm\sqrt{5})}{2} \\\\ x=\dfrac{\cancel{2}(-1\pm\sqrt{5})}{\cancel{2}} \\\\ x=-1\pm\sqrt{5} .\end{array} Hence, the $x-$intercepts are \begin{array}{l}\require{cancel} \text{$x-$intercepts: } \{ -1-\sqrt{5},-1+\sqrt{5} \} .\end{array}
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.