If `x_(1)+ x_(2)=3, x_(3)+x_(4)=12` and `x_(1), x_(2), x_(3), x_(4)` are in increasing G.P then the equation having `x_(1), x_(2)` as roots is

Options

A. `x^(2)-3x+2=0`
B. `x^(2)- 6x+8=0`
C. `x^(2)-12x+32=0`
D. `x^(2)-9x+8=0`

Answer

To solve the problem, we need to find the equation having roots \( x_1 \) and \( x_2 \) given that \( x_1 + x_2 = 3 \), \( x_3 + x_4 = 12 \), and \( x_1, x_2, x_3, x_4 \) are in increasing geometric progression (G.P.).

### Step-by-step Solution:

1. **Define Terms of G.P.:**
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). Then, we can express the terms as:
- \( x_1 = a \)
- \( x_2 = ar \)
- \( x_3 = ar^2 \)
- \( x_4 = ar^3 \)

2. **Set Up Equations:**
From the problem, we have the following equations:
- From \( x_1 + x_2 = 3 \):
\[ a + ar = 3 \quad \text{(Equation 1)} \]
- From \( x_3 + x_4 = 12 \):
\[ ar^2 + ar^3 = 12 \quad \text{(Equation 2)} \]

3. **Simplify Equations:**
- From Equation 1:
\[ a(1 + r) = 3 \quad \Rightarrow \quad a = \frac{3}{1 + r} \quad \text{(Equation 3)} \]
- From Equation 2:
\[ ar^2(1 + r) = 12 \quad \Rightarrow \quad a = \frac{12}{r^2(1 + r)} \quad \text{(Equation 4)} \]

4. **Equate Equations for \( a \):**
Set Equation 3 equal to Equation 4:
\[ \frac{3}{1 + r} = \frac{12}{r^2(1 + r)} \]
Cross-multiplying gives:
\[ 3r^2 = 12 \quad \Rightarrow \quad r^2 = 4 \quad \Rightarrow \quad r = 2 \text{ (since \( r \) must be positive)} \]

5. **Find \( a \):**
Substitute \( r = 2 \) back into Equation 3 to find \( a \):
\[ a = \frac{3}{1 + 2} = \frac{3}{3} = 1 \]

6. **Determine \( x_1 \) and \( x_2 \):**
Now we can find \( x_1 \) and \( x_2 \):
- \( x_1 = a = 1 \)
- \( x_2 = ar = 1 \cdot 2 = 2 \)

7. **Form the Quadratic Equation:**
The quadratic equation with roots \( x_1 \) and \( x_2 \) can be formed using the formula:
\[ x^2 - (x_1 + x_2)x + x_1x_2 = 0 \]
Substituting \( x_1 + x_2 = 3 \) and \( x_1 x_2 = 1 \cdot 2 = 2 \):
\[ x^2 - 3x + 2 = 0 \]

### Final Answer:
The equation having \( x_1 \) and \( x_2 \) as roots is:
\[ \boxed{x^2 - 3x + 2 = 0} \]