Section 4.4
Quadratic Equations from Fractions
If an equation has fractions in it, the way to get rid of the fractions is to multiply both sides of the equation by the least common denominator. We deal here with more complicated denominators than we have seen in the past.
Examples:
a. Solve for $x$: $\dfrac{3}{x}+\dfrac{3}{x+2}=1$.
Solution:
The least common denominator is $x(x+2)$. When we multiply both sides of the equation by the LCD, we obtain
\begin{align}
x(x+2)\left[\frac{3}{x}+\dfrac{3}{x+2}\right]&=\Big[1\Big]x(x+2)\\
3(x+2)+3x&=1\cdot x(x+2)\\
3x+6+3x&=x^2+2x\\
0&=x^2-4x-6\\
6&=x^2-4x\\
6+4&=x^2-4x+4\\
10&=(x-2)^2\\
\pm \sqrt{10}&=x-2\\
x&=2\pm \sqrt{10}
\end{align}
This equation could have arisen from a problem in which two people take three hours to do a job working together. Working alone, one person takes two hours longer than the other to complete the entire task. Since one of the answers is a negative number, and doesn't make sense in the context of that problem, we would only accept the solution $x=2+\sqrt{10}$ hours.
b. Solve for $r$: $\dfrac{120}{r}-\dfrac{120}{r+5}=\dfrac{1}{3}.$
Solution:
The least common denominator of all the fractions in the problem is $3r(r+5)$. When we multiply both sides of the equation by this LCD, we obtain
\begin{align}
3r(r+5)\left[\frac{120}{r}-\frac{120}{r+5}\right]&=\left[\frac{1}{3}\right] 3r(r+5)\\
3(r+5)\cdot 120-3r\cdot 120&=r(r+5)\\
360r+1800-360r&=r^2+5r\\
0&=r^2+5r-1800\\
0&=(r-40)(r+45)\\
r&=40,\;\;-45
\end{align}
This equation could have arisen from a problem in which a person made a 120-mile trip at a certain rate. If she had gone 5 miles per hour faster, the trip would have taken 20 minutes ($\frac{1}{3}$ of an hour) less time. In that context, only the positive solution makes sense. She went at 40 miles per hour.
Exercises:
Solve each of the following equations for x.\begin{array}{l l}
1. \;\;\frac{3}{x+3}-\frac{1}{x-1}=2\;\;& 2. \;\;\frac{4}{x+1}-\frac{1}{x-2}=3\;\; \\
3. \;\;\frac{x-5}{x-7}-\frac{x-2}{x-4}=-3\;\;& 4. \;\;\frac{3}{x+1}+\frac{2}{x-1}=5\;\; \\
5. \;\;\frac{x+2}{x-2}+\frac{x-3}{x+3}=5\;\;& 6. \;\;\frac{x+2}{x-2}+1=\frac{x+3}{3-x}\;\; \\
7. \;\;\frac{1}{(x-2)(x-3)}-\frac{2}{x-2}=3\;\;& 8. \;\;\frac{5}{x+1}+\frac{2}{x-2}=2\;\; \\
9. \;\;\frac{x+1}{x-1}+\frac{2}{x}=4\;\;& 10. \;\;\frac{2}{x-3}-\frac{6}{x(x-3)}=1\;\; \\
11. \;\;\frac{1}{x+4}+\frac{4}{x(x+4)}=2\;\;& 12. \;\;\frac{3}{x-2}-\frac{6}{{{x}^{2}}-2x}=1\;\; \\
13. \;\;\frac{8}{x+6}+\frac{8}{x-6}=1\;\;& 14. \;\;\frac{x}{x+6}+\frac{2}{x-4}=\frac{5x}{{{x}^{2}}+2x-24}\;\; \\
15. \;\;\frac{x}{x+1}+\frac{5}{x-3}=\frac{-4x}{{{x}^{2}}-2x-3}\;\;& 16. \;\;\frac{4}{x-3}-\frac{4}{x+3}=\frac{3}{x}\;\; \\
17. \;\;\frac{8}{x+2}-\frac{8}{x-2}=\frac{x}{{{x}^{2}}-4}\;\;& 18. \;\;\frac{8}{x}-\frac{8}{x(x+2)}=\frac{6x}{x+2}\;\; \\
19. \;\;\frac{8}{x-3}-\frac{8}{(x-3)(x-2)}=4\;\;& 20. \;\;\frac{1}{x-3}-\frac{1}{(x-3)(x-1)}=\frac{x-2}{x-1}\;\; \\
21. \;\;\frac{6}{x}+\frac{2}{x-2}=1\;\;& 22. \;\;\frac{3}{x-1}+\frac{5}{x+3}=1\;\; \\
23. \;\;\frac{x}{x+1}-\frac{6}{1-x}=0\;\;& 24. \;\;\frac{x}{{{x}^{2}}-5x-6}-\frac{1}{{{x}^{2}}-1}=\frac{2x}{{{x}^{2}}-7x+6}\;\; \\
25. \;\;\frac{2}{{{x}^{2}}-3x-4}+\frac{3}{{{x}^{2}}-16}=\frac{x}{{{x}^{2}}+5x+4}\;\;& 26. \;\;\frac{x}{{{x}^{2}}+x-12}+\frac{1}{{{x}^{2}}-5x+6}=\frac{2x}{{{x}^{2}}+2x-8}\;\;
\end{array}