Two non-integer roots of
`((3x -1)/( 2x+3))^(4) - 5 ((3x - 1)/(2x + 3))^(2) + 4 = 0" " ` (1) are

Options

A. `- 5//7, - 2//5`
B. `- 2//5, 7//5`
C. `5//7, 7//5`
D. `-2//5, 3//5`

Answer

To solve the equation

\[ \left(\frac{3x - 1}{2x + 3}\right)^{4} - 5 \left(\frac{3x - 1}{2x + 3}\right)^{2} + 4 = 0, \]

we can use a substitution method. Let

\[ t = \left(\frac{3x - 1}{2x + 3}\right)^{2}. \]

This transforms our equation into:

\[ t^{2} - 5t + 4 = 0. \]

### Step 1: Factor the quadratic equation

We can factor the quadratic equation:

\[ t^{2} - 5t + 4 = (t - 1)(t - 4) = 0. \]

### Step 2: Find the values of t

Setting each factor to zero gives us:

1. \(t - 1 = 0 \Rightarrow t = 1\)
2. \(t - 4 = 0 \Rightarrow t = 4\)

### Step 3: Substitute back to find x

Now we substitute back for \(t\) in terms of \(x\):

1. For \(t = 1\):

\[ \left(\frac{3x - 1}{2x + 3}\right)^{2} = 1 \Rightarrow \frac{3x - 1}{2x + 3} = \pm 1. \]

- Case 1:

\[ \frac{3x - 1}{2x + 3} = 1 \Rightarrow 3x - 1 = 2x + 3 \Rightarrow x = 4. \]

- Case 2:

\[ \frac{3x - 1}{2x + 3} = -1 \Rightarrow 3x - 1 = -2x - 3 \Rightarrow 5x = -2 \Rightarrow x = -\frac{2}{5}. \]

2. For \(t = 4\):

\[ \left(\frac{3x - 1}{2x + 3}\right)^{2} = 4 \Rightarrow \frac{3x - 1}{2x + 3} = \pm 2. \]

- Case 1:

\[ \frac{3x - 1}{2x + 3} = 2 \Rightarrow 3x - 1 = 4x + 6 \Rightarrow -x = 7 \Rightarrow x = -7. \]

- Case 2:

\[ \frac{3x - 1}{2x + 3} = -2 \Rightarrow 3x - 1 = -4x - 6 \Rightarrow 7x = -5 \Rightarrow x = -\frac{5}{7}. \]

### Step 4: Collect the non-integer roots

The solutions we found are:

1. \(x = 4\) (integer)
2. \(x = -\frac{2}{5}\) (non-integer)
3. \(x = -7\) (integer)
4. \(x = -\frac{5}{7}\) (non-integer)

Thus, the two non-integer roots are:

\[ x = -\frac{2}{5} \quad \text{and} \quad x = -\frac{5}{7}. \]