Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set: 31

Answer

The solutions are $\dfrac{-9- \sqrt{201}}{6}$ and $\dfrac{-9+ \sqrt{201}}{6}$.

Work Step by Step

Multiply the LCD $(x-2)(x+3)$ to both sides of the equation to get rid of the denominators: \begin{array}{ccc} &(x-2)(x+3)\left[\dfrac{2x}{x-2}+\dfrac{x}{x+3}\right] &= &-\dfrac{5}{x+3} \cdot (x-2)(x+3) \\&(x-2)(x+3) \cdot \dfrac{2x}{x-2}+(x-2)(x+3) \cdot \dfrac{x}{x+3} &= &-\dfrac{5}{x+3} \cdot (x-2)(x+3) \end{array} Cancel simplify by cancelling common factors to obtain: \begin{array}{ccc} \require{cancel} \\&\cancel{(x-2)}(x+3) \cdot \dfrac{2x}{\cancel{x-2}}+(x-2)\cancel{(x+3)} \cdot \dfrac{x}{\cancel{x+3}} &= &-\dfrac{5}{\cancel{x+3}} \cdot (x-2)\cancel{(x+3)} \\&(x+3)(2x)+(x-2)(x) &= &-5(x-2) \\&2x(x) + 2x(3) + x(x)-x(2) &= &-5(x) -(-5)(2) \\&2x^2+6x+x^2-2x &= &-5x+10 \\&3x^2+4x &= &-5x+10 \end{array} Put all terms on the left side. Note that when a term is transferred to the other side of an equation, it changes it sign to its opposite. \begin{array}{ccc} \\&3x^2+4x &= &-5x+10 \\&3x^2+4x+5x-10 &= &0 \\&3x^2+9x-10 &= &0 \end{array} The trinomial cannot be factored. Solve the equation using the quadratic formula with $a=3, b=9, c=-10$ to obtain: $x=\dfrac{-9\pm \sqrt{9^2-4(3)(-10)}}{2(3)} \\x=\dfrac{-9\pm\sqrt{81+120}}{6} \\x=\dfrac{-9\pm \sqrt{201}}{6}$ Therefore, the solutions are $\dfrac{-9- \sqrt{201}}{6}$ and $\dfrac{-9+ \sqrt{201}}{6}$.
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