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: 19

Answer

$x_{1}= \dfrac{4}{3}$ and $x_{2} = -2$

Work Step by Step

Given $3x^2+2x=8 \longrightarrow 3x^2+2x-8=0$ $a= 3, \ b=2, \ c=-8$ Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ we have: $\dfrac{-2 \pm \sqrt{2^2-4 \times 3\times (-8)}}{2 \times 3} = \dfrac{-2 \pm \sqrt{4+96}}{6} = \dfrac{-2 \pm \sqrt{100}}{6} = \dfrac{-2 \pm 10}{6} = \dfrac{-1 \pm 5}{3}$ Therefore the solutions are $x_{1}= \dfrac{-1 + 5}{3} =\dfrac{4}{3}$ and $x_{2} = \dfrac{-1 - 5}{3} = \dfrac{-6}{3} = -2$
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