Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set: 5

Answer

{$-4,1$}

Work Step by Step

First, we subtract the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $n$: $n-\frac{4}{n}=-3$ $\frac{n(n)-4(1)}{n}=-3$ $\frac{n^{2}-4}{n}=-3$ $\frac{n^{2}-4}{n}=-\frac{3}{1}$ Now, we cross multiply the two fractions in order to create a quadratic equation: $\frac{n^{2}-4}{n}=-\frac{3}{1}$ $1(n^{2}-4)=-3n$ $n^{2}-4=-3n$ $n^{2}+3n-4=0$ Now, we use rules of factoring trinomials to solve the equation: $n^{2}+3n-4=0$ $n^{2}-1n+4n-4=0$ $n(n-1)+4(n-1)=0$ $(n-1)(n+4)=0$ $(n-1)=0$ or $(n+4)=0$ $n=1$ or $n=-4$ Therefore, the solution is {$-4,1$}.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.