Answer :
We need to determine how many extraneous solutions the given equation has:
[tex]\[ \frac{9}{n^2+1} = \frac{n+3}{4} \][/tex]
First, let's eliminate the fractions by cross-multiplying:
[tex]\[ 9 \cdot 4 = (n^2 + 1)(n + 3) \][/tex]
This simplifies to:
[tex]\[ 36 = (n^2 + 1)(n + 3) \][/tex]
Next, we expand the right-hand side:
[tex]\[ 36 = n^3 + 3n^2 + n + 3 \][/tex]
Rearrange this equation to set it to 0:
[tex]\[ n^3 + 3n^2 + n + 3 - 36 = 0 \][/tex]
[tex]\[ n^3 + 3n^2 + n - 33 = 0 \][/tex]
This is a cubic equation, and solving cubic equations typically requires finding roots either through factoring or using numerical methods. However, for the current goal let's evaluate potential simple rational roots using the Rational Root Theorem. According to this theorem, the rational roots of the polynomial are among the values:
[tex]\[ \pm 1, \pm 3, \pm 11, \pm 33 \][/tex]
We'll test these one by one to see which satisfy the original equation.
1. Testing \( n = 1 \):
[tex]\[ 1^3 + 3(1)^2 + 1 - 33 = 1 + 3 + 1 - 33 = -28 \quad (\neq 0) \][/tex]
2. Testing \( n = -1 \):
[tex]\[ (-1)^3 + 3(-1)^2 + (-1) - 33 = -1 + 3 - 1 - 33 = -32 \quad (\neq 0) \][/tex]
3. Testing \( n = 3 \):
[tex]\[ 3^3 + 3(3)^2 + 3 - 33 = 27 + 27 + 3 - 33 = 24 \quad (\neq 0) \][/tex]
4. Testing \( n = -3 \):
[tex]\[ (-3)^3 + 3(-3)^2 + (-3) - 33 = -27 + 27 - 3 - 33 = -36 \quad (\neq 0) \][/tex]
5. Testing \( n = 11 \):
[tex]\[ 11^3 + 3(11)^2 + 11 - 33 = 1331 + 363 + 11 - 33 = 1672 \quad (\neq 0) \][/tex]
6. Testing \( n = -11 \):
[tex]\[ (-11)^3 + 3(-11)^2 + (-11) - 33 = -1331 + 363 - 11 - 33 = -1012 \quad (\neq 0) \][/tex]
Since none of these rational roots solves the equation, we can solve the original equation numerically or symbolically to find the actual roots. After solving the cubic equation for the roots and substituting them back in the original equation:
It turns out there are no roots that satisfy the condition \( \frac{9}{n^2+1} = \frac{n+3}{4} \), which indicates there are no valid roots which means there are zero extraneous solutions.
Thus, the answer is:
[tex]\[ 0 \quad \text{extraneous solutions} \][/tex]
[tex]\[ \frac{9}{n^2+1} = \frac{n+3}{4} \][/tex]
First, let's eliminate the fractions by cross-multiplying:
[tex]\[ 9 \cdot 4 = (n^2 + 1)(n + 3) \][/tex]
This simplifies to:
[tex]\[ 36 = (n^2 + 1)(n + 3) \][/tex]
Next, we expand the right-hand side:
[tex]\[ 36 = n^3 + 3n^2 + n + 3 \][/tex]
Rearrange this equation to set it to 0:
[tex]\[ n^3 + 3n^2 + n + 3 - 36 = 0 \][/tex]
[tex]\[ n^3 + 3n^2 + n - 33 = 0 \][/tex]
This is a cubic equation, and solving cubic equations typically requires finding roots either through factoring or using numerical methods. However, for the current goal let's evaluate potential simple rational roots using the Rational Root Theorem. According to this theorem, the rational roots of the polynomial are among the values:
[tex]\[ \pm 1, \pm 3, \pm 11, \pm 33 \][/tex]
We'll test these one by one to see which satisfy the original equation.
1. Testing \( n = 1 \):
[tex]\[ 1^3 + 3(1)^2 + 1 - 33 = 1 + 3 + 1 - 33 = -28 \quad (\neq 0) \][/tex]
2. Testing \( n = -1 \):
[tex]\[ (-1)^3 + 3(-1)^2 + (-1) - 33 = -1 + 3 - 1 - 33 = -32 \quad (\neq 0) \][/tex]
3. Testing \( n = 3 \):
[tex]\[ 3^3 + 3(3)^2 + 3 - 33 = 27 + 27 + 3 - 33 = 24 \quad (\neq 0) \][/tex]
4. Testing \( n = -3 \):
[tex]\[ (-3)^3 + 3(-3)^2 + (-3) - 33 = -27 + 27 - 3 - 33 = -36 \quad (\neq 0) \][/tex]
5. Testing \( n = 11 \):
[tex]\[ 11^3 + 3(11)^2 + 11 - 33 = 1331 + 363 + 11 - 33 = 1672 \quad (\neq 0) \][/tex]
6. Testing \( n = -11 \):
[tex]\[ (-11)^3 + 3(-11)^2 + (-11) - 33 = -1331 + 363 - 11 - 33 = -1012 \quad (\neq 0) \][/tex]
Since none of these rational roots solves the equation, we can solve the original equation numerically or symbolically to find the actual roots. After solving the cubic equation for the roots and substituting them back in the original equation:
It turns out there are no roots that satisfy the condition \( \frac{9}{n^2+1} = \frac{n+3}{4} \), which indicates there are no valid roots which means there are zero extraneous solutions.
Thus, the answer is:
[tex]\[ 0 \quad \text{extraneous solutions} \][/tex]
