College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 66


$\displaystyle x=\frac{13\pm\sqrt{145}}{4}$

Work Step by Step

We are given: $\frac{2x}{x-3}+\frac{1}{x}=4$ We multiply through by $x(x-3)$: $x(x-3)(\frac{2x}{x-3}+\frac{1}{x})=x(x-3)*4$ $2x*x+(x-3)=4x^{2}-12x$ $2x^{2}+x-3=4x^{2}-12x$ $2x^{2}-13x+3=0$ We solve using the quadratic formula ($a=2,\ b=-13,\ c=3$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $\displaystyle x=\frac{-(-13)\pm\sqrt{(-13)^{2}-4*2*3}}{2*2}$ $\displaystyle x=\frac{-(-13)\pm\sqrt{169-24}}{4}$ $\displaystyle x=\frac{13\pm\sqrt{145}}{4}$
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