Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Review: 58

Answer

$x_{1} = \dfrac{4}{5}$ and $x_{2}= -\dfrac{1}{2}$

Work Step by Step

Given $10x^2=3x+4 \longrightarrow 10x^2-3x-4=0$ $a=10, \ b=-3, \ c=-4$ Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} ,$ we have: $\dfrac{-(-3)\pm \sqrt{(-3)^2-4\times 10\times (-4)}}{2\times 10} = \dfrac{3\pm \sqrt{9+160}}{20} = \dfrac{3\pm \sqrt{169}}{20} = \dfrac{3\pm 13}{20}$ Therefore, the solutions are $x_{1}= \dfrac{3 + 13}{20} = \dfrac{16}{20} = \dfrac{4}{5}$ and $x_{2}= \dfrac{3- 13}{20} = \dfrac{-10}{20} = -\dfrac{1}{2}$
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