EQUATIONS INVOLVING FRACTIONS (RATIONAL EQUATIONS)

Note:



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Solve for x in the following equation.


Problem 5.5 c:         $\displaystyle \frac{500}{2x^{2}-9x+1}+1=x-5$



Answer:     $x\approx 10.23$



Simplify the equation by subtracting 1 from both sides of the equation..

\begin{eqnarray*}&& \\
\displaystyle \frac{500}{2x^{2}-9x+1}+1 &=&x-5 \\
&& \\...
...displaystyle \frac{500}{2x^{2}-9x+1} &=&x-6 \\
&& \\
&& \\
&&
\end{eqnarray*}




Multiply both sides of the equation by an expression that represents the lowest common denominator. The expression $\left( 2x^{2}-9x+1\right) $ is the smallest expression because it is the smallest expression that is divisible by all the denominators.




\begin{eqnarray*}&& \\
\displaystyle \frac{500}{2x^{2}-9x+1} &=&x-6 \\
&& \\
...
...
&& \\
&& \\
0 &=&2x^{3}-21x^{2}+55x-506 \\
&& \\
&& \\
&&
\end{eqnarray*}


This expression is not easily factored. You can solve this equation using iteration or by graphing.



With iteration, choose two real numbers and evaluate f(x)=2x3-21x2+55x-506 at each of the two numbers. You are looking for a sign change. If there is a sign change, then you know that one real solution is located between the two values you choose to test.

\begin{eqnarray*}&& \\
f(0) &=&-506 \\
&& \\
f\left( 20\right) &=&8194 \\
&& \\
&&
\end{eqnarray*}


This indicates that a real solution exists between 0 and 20. Now narrow the range to 10 and 20.

\begin{eqnarray*}&& \\
f(10) &=&-56 \\
&& \\
f\left( 20\right) &=&8194 \\
&&
\end{eqnarray*}


This indicates that real solution exists between 10 and 20. Narrow the range again. Since -56 is close to 0, choose a number near 10.

\begin{eqnarray*}&& \\
f\left( 11\right) &=&220 \\
&& \\
f\left( 10\right) &=&-56 \\
&&
\end{eqnarray*}


You know a real solution exists between 10 and 11. Narrow the range again.

\begin{eqnarray*}&& \\
f\left( 10.5\right) &=&71.5 \\
&& \\
f\left( 10\right) &=&-56 \\
&&
\end{eqnarray*}


You know a real solution exists between 10 and 1.5. Narrow the range again.

\begin{eqnarray*}&& \\
f\left( 10.25\right) &=&5.219 \\
&& \\
f\left( 10\right) &=&-56 \\
&&
\end{eqnarray*}


You know a real solution exists between 10 and 10.25. Narrow the range again.

\begin{eqnarray*}&& \\
f\left( 10.2\right) &=&-7.424 \\
&& \\
f\left( 10.25\right) &=&5.219 \\
&&
\end{eqnarray*}


You know a real solution exists between 10.2 and 10.25. You can keep going until you have the accuracy you need in your answer.






You can also check your answer by graphing

\begin{eqnarray*}&& \\
f(x) &=&\displaystyle \frac{500}{2x^{2}-9x+1}-x+6 \\
&&
\end{eqnarray*}


$.\quad $(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis at one spot, $
10.2294572878\approx 10.23$






If you would like to test yourself by working some problems similar to this example, click on problem.




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Author: Nancy Marcus

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