The number of roots of `81 ((2x - 5)/( 3x +1))^(4) - 45 ((2x - 5)/(3x + 1))^(2) + 4 = 0 , x ne 1//3` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( 81 \left( \frac{2x - 5}{3x + 1} \right)^4 - 45 \left( \frac{2x - 5}{3x + 1} \right)^2 + 4 = 0 \) for the number of roots, we can follow these steps:

### Step 1: Substitute \( y = \left( \frac{2x - 5}{3x + 1} \right)^2 \)

This substitution simplifies the equation into a quadratic form:
\[ 81y^2 - 45y + 4 = 0 \]

### Step 2: Identify the coefficients

From the quadratic equation \( 81y^2 - 45y + 4 = 0 \), we have:
- \( a = 81 \)
- \( b = -45 \)
- \( c = 4 \)

### Step 3: Calculate the discriminant

The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = (-45)^2 - 4 \cdot 81 \cdot 4 \]
\[ D = 2025 - 1296 \]
\[ D = 729 \]

### Step 4: Determine the nature of the roots

Since the discriminant \( D = 729 \) is positive, the quadratic equation has two distinct real roots.

### Step 5: Find the values of \( y \)

Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ y = \frac{45 \pm \sqrt{729}}{2 \cdot 81} \]
\[ y = \frac{45 \pm 27}{162} \]
Calculating the two possible values for \( y \):
1. \( y_1 = \frac{72}{162} = \frac{4}{9} \)
2. \( y_2 = \frac{18}{162} = \frac{1}{9} \)

### Step 6: Solve for \( x \)

Recall that \( y = \left( \frac{2x - 5}{3x + 1} \right)^2 \). We will solve for \( x \) for each value of \( y \).

#### For \( y_1 = \frac{4}{9} \):
\[ \frac{2x - 5}{3x + 1} = \pm \frac{2}{3} \]

1. **Case 1**: \( \frac{2x - 5}{3x + 1} = \frac{2}{3} \)
\[ 3(2x - 5) = 2(3x + 1) \]
\[ 6x - 15 = 6x + 2 \implies -15 = 2 \quad \text{(no solution)} \]

2. **Case 2**: \( \frac{2x - 5}{3x + 1} = -\frac{2}{3} \)
\[ 3(2x - 5) = -2(3x + 1) \]
\[ 6x - 15 = -6x - 2 \]
\[ 12x = 13 \implies x = \frac{13}{12} \]

#### For \( y_2 = \frac{1}{9} \):
\[ \frac{2x - 5}{3x + 1} = \pm \frac{1}{3} \]

1. **Case 1**: \( \frac{2x - 5}{3x + 1} = \frac{1}{3} \)
\[ 3(2x - 5) = 1(3x + 1) \]
\[ 6x - 15 = 3x + 1 \]
\[ 3x = 16 \implies x = \frac{16}{3} \]

2. **Case 2**: \( \frac{2x - 5}{3x + 1} = -\frac{1}{3} \)
\[ 3(2x - 5) = -1(3x + 1) \]
\[ 6x - 15 = -3x - 1 \]
\[ 9x = 14 \implies x = \frac{14}{9} \]

### Conclusion

The values of \( x \) obtained are \( \frac{13}{12}, \frac{16}{3}, \frac{14}{9} \). Since all these values are distinct and valid, the total number of roots of the original equation is **3**.