how many extraneous solutions does the equation below have?\n$\frac{9}{n^{2}+1}=\frac{n + 3}{4}$\n0\n1\n2\n3
how many extraneous solutions does the equation below have?\n$\frac{9}{n^{2}+1}=\frac{n + 3}{4}$\n0\n1\n2\n3
Answer
Answer:
A. 0
Explanation:
Step1: Cross - multiply
$9\times4=(n + 3)(n^{2}+1)$ $36=n^{3}+n+3n^{2}+3$ $n^{3}+3n^{2}+n - 33 = 0$. The domain of the original rational equation $\frac{9}{n^{2}+1}=\frac{n + 3}{4}$ is all real numbers since $n^{2}+1\neq0$ for all real $n$. When we solve the resulting polynomial equation $n^{3}+3n^{2}+n - 33 = 0$, any solutions of this polynomial equation are valid solutions of the original rational equation. So, there are 0 extraneous solutions.