Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Cumulative Review: 40

Answer

$x_{1}=2+ 2\sqrt{3}$ and $x_{2}=2- 2\sqrt{3}$

Work Step by Step

Given $m^2=4m+8 \longrightarrow m^2-4m-8=0$ $a=1, \ b=-4, \ c=-8$ Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have: $\dfrac{-(-4)\pm \sqrt{(-4)^2-4\times 1\times (-8)}}{2\times 1} = \dfrac{4\pm \sqrt{16+32}}{2} = \dfrac{4\pm \sqrt{48}}{2} = \dfrac{4\pm \sqrt{4^2\times 3}}{2} = \dfrac{4\pm 4\sqrt{3}}{2} = 2\pm 2\sqrt{3}$ Therefore the solutions are $x_{1}=2+ 2\sqrt{3}$ and $x_{2}=2- 2\sqrt{3}$
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