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: 11

Answer

$x=\dfrac{1}{2}\pm\dfrac{\sqrt{29}}{2}$

Work Step by Step

$\dfrac{7}{x^{2}-5x+6}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}$ Factor the denominator of the fraction on the left side of the equation: $\dfrac{7}{(x-2)(x-3)}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}$ Multiply the whole equation by $(x-2)(x-3)$: $(x-2)(x-3)\Big[\dfrac{7}{(x-2)(x-3)}=\dfrac{2x}{x-3}-\dfrac{x}{x-2}\Big]$ $7=2x(x-2)-x(x-3)$ $7=2x^{2}-4x-x^{2}+3x$ Take all terms to the right side of the equation and simplify it: $0=2x^{2}-4x-x^{2}+3x-7$ $x^{2}-x-7=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=1$, $b=-1$ and $c=-7$. Substitute the known values into the formula and simplify: $x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-7)}}{2(1)}=\dfrac{1\pm\sqrt{1+28}}{2}=...$ $...=\dfrac{1\pm\sqrt{29}}{2}=\dfrac{1}{2}\pm\dfrac{\sqrt{29}}{2}$ The original equation is not undefined for neither of the values of $x$ found. Our final answer is $x=\dfrac{1}{2}\pm\dfrac{\sqrt{29}}{2}$
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