Question

how many extraneous solutions does the equation have?\\(\frac{9}{n^2 + 1} = \frac{n + 3}{4}\\)\
1\
2\
3\
0

Explanation:

Step1: Analyze the denominator

The denominator of the left - hand side is \(n^{2}+1\). For any real number \(n\), \(n^{2}\geqslant0\), so \(n^{2} + 1\geqslant1\). This means that the denominator \(n^{2}+1\) is never equal to zero for real values of \(n\).

Step2: Solve the equation

Cross - multiply the equation \(\frac{9}{n^{2}+1}=\frac{n + 3}{4}\) to get \(9\times4=(n + 3)(n^{2}+1)\).
Expand the right - hand side: \(36=n^{3}+n+3n^{2}+3\).
Rearrange the equation to the standard cubic form: \(n^{3}+3n^{2}+n - 33 = 0\).
We can try to find real roots of this cubic equation. By the Rational Root Theorem, the possible rational roots are factors of \(33\) divided by factors of \(1\), i.e., \(\pm1,\pm3,\pm11,\pm33\).

• When \(n = 3\): \(3^{3}+3\times3^{2}+3 - 33=27 + 27+3 - 33=24

eq0\).

• When \(n=-3\): \((-3)^{3}+3\times(-3)^{2}+(-3)-33=-27 + 27-3 - 33=-36

eq0\).

• When \(n = 1\): \(1^{3}+3\times1^{2}+1 - 33=1 + 3+1 - 33=-28

eq0\).

• When \(n=-1\): \((-1)^{3}+3\times(-1)^{2}+(-1)-33=-1 + 3-1 - 33=-32

eq0\).

• When \(n = 11\): \(11^{3}+3\times11^{2}+11 - 33=1331+363 + 11-33 = 1672

eq0\).

• When \(n=-11\): \((-11)^{3}+3\times(-11)^{2}+(-11)-33=-1331 + 363-11 - 33=-1012

eq0\).

• When \(n = 33\): \(33^{3}+3\times33^{2}+33 - 33\) is a large positive number, not zero.
• When \(n=-33\): \((-33)^{3}+3\times(-33)^{2}+(-33)-33\) is a large negative number, not zero.

We can also use the fact that since the denominator \(n^{2}+1\) is never zero for real \(n\), any solution of the original rational equation (after cross - multiplying) will be a valid solution of the original equation because there are no values of \(n\) that make the original denominator zero. So, there are no extraneous solutions.

Answer:

0 (the option with the value 0)