x and y are positive number.Let g and a be G.M. And AM of these numbers.Also let G be G.M of x + 1 and y + 1.If G and g are roots of equation `x^(2) - 5x + 6 = 0` , then

Options

A. `x = 2, y = (3)/(4)`
B. `x = (3)/(4), y = 12`
C. `x = (5)/(2), y = (8)/(5)`
D. x = y = 2

Answer

To solve the problem step by step, we will follow the given information and derive the necessary equations.

### Step 1: Understand the Given Information
We have two positive numbers \( x \) and \( y \). The geometric mean (G.M.) \( g \) and the arithmetic mean (A.M.) \( a \) of these numbers are defined as follows:
- \( g = \sqrt{xy} \)
- \( a = \frac{x + y}{2} \)

We also have another G.M. \( G \) defined for \( x + 1 \) and \( y + 1 \):
- \( G = \sqrt{(x + 1)(y + 1)} \)

### Step 2: Roots of the Quadratic Equation
The roots of the equation \( x^2 - 5x + 6 = 0 \) can be found using the factorization method:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \]
Thus, the roots are \( x = 2 \) and \( x = 3 \). We can assign:
- \( g = 2 \)
- \( G = 3 \)

### Step 3: Set Up Equations
From the definitions of \( g \) and \( G \), we can set up the following equations:
1. \( g = \sqrt{xy} = 2 \)
2. \( G = \sqrt{(x + 1)(y + 1)} = 3 \)

### Step 4: Solve for \( xy \)
From the first equation:
\[ \sqrt{xy} = 2 \implies xy = 2^2 = 4 \]

### Step 5: Solve for \( (x + 1)(y + 1) \)
From the second equation:
\[ \sqrt{(x + 1)(y + 1)} = 3 \implies (x + 1)(y + 1) = 3^2 = 9 \]
Expanding this, we have:
\[ xy + x + y + 1 = 9 \]
Substituting \( xy = 4 \):
\[ 4 + x + y + 1 = 9 \implies x + y + 5 = 9 \implies x + y = 4 \]

### Step 6: Solve the System of Equations
Now we have a system of equations:
1. \( xy = 4 \)
2. \( x + y = 4 \)

Let \( x \) and \( y \) be the roots of the quadratic equation \( t^2 - (x+y)t + xy = 0 \):
\[ t^2 - 4t + 4 = 0 \]
Factoring gives:
\[ (t - 2)^2 = 0 \]
Thus, \( t = 2 \) is a double root, meaning:
\[ x = 2 \quad \text{and} \quad y = 2 \]

### Final Answer
The values of \( x \) and \( y \) are both \( 2 \).

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