The number of real roots of `((x - 1)/( x + 1))^(4) - 13 ((x - 1)/( x + 1))^(2) + 36 = 0, x ne - 1` is

Options

A. 0
B. 2
C. 3
D. 4

Answer

To find the number of real roots of the equation

\[ \left(\frac{x - 1}{x + 1}\right)^{4} - 13 \left(\frac{x - 1}{x + 1}\right)^{2} + 36 = 0, \quad x \neq -1, \]

we can start by substituting \( y = \left(\frac{x - 1}{x + 1}\right)^{2} \). This substitution simplifies our equation to a quadratic form:

\[ y^{2} - 13y + 36 = 0. \]

### Step 1: Solve the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}, \]

where \( a = 1, b = -13, c = 36 \).

Substituting these values into the formula gives:

\[ y = \frac{13 \pm \sqrt{(-13)^{2} - 4 \cdot 1 \cdot 36}}{2 \cdot 1}. \]

Calculating the discriminant:

\[ (-13)^{2} = 169, \]
\[ 4 \cdot 1 \cdot 36 = 144, \]
\[ b^{2} - 4ac = 169 - 144 = 25. \]

Now substituting back into the formula:

\[ y = \frac{13 \pm \sqrt{25}}{2} = \frac{13 \pm 5}{2}. \]

Calculating the two possible values for \( y \):

1. \( y = \frac{13 + 5}{2} = \frac{18}{2} = 9 \)
2. \( y = \frac{13 - 5}{2} = \frac{8}{2} = 4 \)

### Step 2: Find the Corresponding Values of \( x \)

Now we have two values for \( y \): \( y = 9 \) and \( y = 4 \).

Recall that:

\[ y = \left(\frac{x - 1}{x + 1}\right)^{2}. \]

#### Case 1: \( y = 9 \)

Taking the square root, we have:

\[ \frac{x - 1}{x + 1} = \pm 3. \]

**Subcase 1.1:** \( \frac{x - 1}{x + 1} = 3 \)

Cross-multiplying gives:

\[ x - 1 = 3(x + 1) \implies x - 1 = 3x + 3 \implies -2x = 4 \implies x = -2. \]

**Subcase 1.2:** \( \frac{x - 1}{x + 1} = -3 \)

Cross-multiplying gives:

\[ x - 1 = -3(x + 1) \implies x - 1 = -3x - 3 \implies 4x = -2 \implies x = -\frac{1}{2}. \]

#### Case 2: \( y = 4 \)

Taking the square root, we have:

\[ \frac{x - 1}{x + 1} = \pm 2. \]

**Subcase 2.1:** \( \frac{x - 1}{x + 1} = 2 \)

Cross-multiplying gives:

\[ x - 1 = 2(x + 1) \implies x - 1 = 2x + 2 \implies -x = 3 \implies x = -3. \]

**Subcase 2.2:** \( \frac{x - 1}{x + 1} = -2 \)

Cross-multiplying gives:

\[ x - 1 = -2(x + 1) \implies x - 1 = -2x - 2 \implies 3x = -1 \implies x = -\frac{1}{3}. \]

### Step 3: Summary of Roots

The values of \( x \) we found are:

1. \( x = -2 \)
2. \( x = -\frac{1}{2} \)
3. \( x = -3 \)
4. \( x = -\frac{1}{3} \)

### Conclusion

Thus, the total number of real roots of the original equation is **4**.

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