Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.4 - Solving Quadratic Equations - Which Method? - Problem Set 10.4: 33

Answer

{$\frac{3}{4},\frac{4}{3}$}

Work Step by Step

First, we add the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $x$: $x+\frac{1}{x}=\frac{25}{12}$ $\frac{x(x)+1(1)}{x}=\frac{25}{12}$ $\frac{x^{2}+1}{x}=\frac{25}{12}$ Now, we cross multiply the two fractions in order to create a quadratic equation: $\frac{x^{2}+1}{x}=\frac{25}{12}$ $12(x^{2}+1)=25(x)$ $12x^{2}+12=25x$ $12x^{2}-25x+12=0$ Now, we use rules of factoring trinomials to solve the equation: $12x^{2}-25x+12=0$ $12x^{2}-9x-16x+12=0$ $3x(4x-3)-4(4x-3)=0$ $(4x-3)(3x-4)=0$ $(4x-3)=0$ or $(3x-4)=0$ $x=\frac{3}{4}$ or $x=\frac{4}{3}$ Therefore, the solution is {$\frac{3}{4},\frac{4}{3}$}.
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