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: 10

Answer

$\text{$x-$intercepts: } \{ -5.4,-2.6 \} $

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+8x+14 ,$ is/are \begin{array}{l}\require{cancel} x^2+8x+14=0 \\\\ x=\dfrac{-8\pm\sqrt{8^2-4(1)(14)}}{2(1)} \\\\ x=\dfrac{-8\pm\sqrt{64-56}}{2} \\\\ x=\dfrac{-8\pm\sqrt{8}}{2} \\\\ x=\dfrac{-8\pm\sqrt{4\cdot2}}{2} \\\\ x=\dfrac{-8\pm\sqrt{(2)^2\cdot2}}{2} \\\\ x=\dfrac{-8\pm2\sqrt{2}}{2} \\\\ x=\dfrac{2(-4\pm\sqrt{2})}{2} \\\\ x=\dfrac{\cancel{2}(-4\pm\sqrt{2})}{\cancel{2}} \\\\ x=-4\pm\sqrt{2} .\end{array} Hence, the $x-$intercepts are \begin{array}{l}\require{cancel} \text{$x-$intercepts: } \{ -5.4,-2.6 \} .\end{array}
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