Chapter 11.8, Problem 3CYP

To determine

To solve and check: the given function

Answer to Problem 3CYP

The required solution is $n=-1$

Explanation of Solution

Given:

Calculation:

  $\begin{array}{l}\frac{n^2-3n}{n^2-4}-\frac{10}{n^2-4}=2\\ \Rightarrow \frac{n^2-3n-10}{n^2-4}=2\\ \Rightarrow \frac{\left(n-5\right)\left(n+2\right)}{\left(n+2\right)\left(n-2\right)}=2\\ \Rightarrow \frac{n-5}{n-2}=2\\ \Rightarrow n-5=2n-4\\ \Rightarrow -5+4=2n-n\\ \Rightarrow n=-1\end{array}$

Check:

  $\begin{array}{l}\frac{{\left(-1\right)}^2-3\left(-1\right)}{{\left(-1\right)}^2-4}-\frac{10}{{\left(-1\right)}^2-4}=2\\ \Rightarrow \frac{1+3}{1-4}-\frac{10}{1-4}=2\\ \Rightarrow \frac{4}{-3}-\frac{10}{-3}=2\\ \Rightarrow \frac{4-10}{-3}=2\\ \Rightarrow \frac{-6}{-3}=2\\ \Rightarrow 2=2\end{array}$

Conclusion:

Therefore, the required solution is $n=-1$