how many extraneous solutions does the equation below have?\n\\(\\frac{9}{n^{2}+1}=\\frac{n +…

Explanation:

Step1: Cross - multiply the equation

$9\times4=(n + 3)(n^{2}+1)$ $36=n^{3}+n+3n^{2}+3$ $n^{3}+3n^{2}+n - 33 = 0$ The denominator $n^{2}+1$ is never zero for real - valued $n$ since $n^{2}\geq0$ for all real $n$, so $n^{2}+1\geq1$ for all real $n$.

Step2: Analyze extraneous solutions

Since the original rational equation has a non - zero denominator for all real $n$ (the denominator $n^{2}+1$ is non - zero for all real $n$), there are no values of $n$ that make the original equation undefined when substituting the solutions of the resulting polynomial equation back into the original equation. So, there are 0 extraneous solutions.

Answer:

A. 0