Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set: 12



Work Step by Step

$\dfrac{11}{2x^{2}+x-15}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}$ Factor the denominator of the fraction on the left side of the equation: $\dfrac{11}{(2x-5)(x+3)}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}$ Multiply the whole equation by $(2x-5)(x+3)$: $(2x-5)(x+3)\Big[\dfrac{11}{(2x-5)(x+3)}=\dfrac{5}{2x-5}-\dfrac{x}{x+3}\Big]$ $11=5(x+3)-x(2x-5)$ $11=5x+15-2x^{2}+5x$ Take all terms to the left side of the equation and simplify it: $11-5x-15+2x^{2}-5x=0$ $2x^{2}-10x-4=0$ Take out common factor $2$ from this equation and simplify it further by taking the common factor to divide the right side of the equation: $2(x^{2}-5x-2)=0$ $x^{2}-5x-2=\dfrac{0}{2}$ $x^{2}-5x-2=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=1$, $b=-5$ and $c=-2$. Substitute the known values into the formula and simplify: $x=\dfrac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(-2)}}{2(1)}=\dfrac{5\pm\sqrt{25+8}}{2}=...$ $...=\dfrac{5\pm\sqrt{33}}{2}=\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$ The original equation is not undefined for neither of the values of $x$ found. Our final answer is $x=\dfrac{5}{2}\pm\dfrac{\sqrt{33}}{2}$
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