If `x_(1)` and `x_(2)` are the arithmetic and harmonic means of the roots fo the equation `ax^(2)+bx+c=0`, the quadratic equation whose roots are `x_(1)` and `x_(2)` is

Options

A. `abx^(2)+(b^(2)+ac)x+bc=0`
B. `2abx^(2)+(b^(2)+4ac)x+2bc=0`
C. `2abx^(2)+(b^(2)+ac)x+bc=0`
D. none of these

Explanation

To solve the problem step by step, we need to find the quadratic equation whose roots are the arithmetic mean \( x_1 \) and the harmonic mean \( x_2 \) of the roots of the given quadratic equation \( ax^2 + bx + c = 0 \).

### Step 1: Identify the roots of the original equation
Let the roots of the equation \( ax^2 + bx + c = 0 \) be \( \alpha \) and \( \beta \).

### Step 2: Calculate the sum and product of the roots
From Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

### Step 3: Find the arithmetic mean \( x_1 \)
The arithmetic mean \( x_1 \) of the roots \( \alpha \) and \( \beta \) is given by:
\[ x_1 = \frac{\alpha + \beta}{2} = \frac{-\frac{b}{a}}{2} = -\frac{b}{2a} \]

### Step 4: Find the harmonic mean \( x_2 \)
The harmonic mean \( x_2 \) of the roots \( \alpha \) and \( \beta \) is given by:
\[ x_2 = \frac{2\alpha\beta}{\alpha + \beta} \]
Substituting the values of \( \alpha + \beta \) and \( \alpha \beta \):
\[ x_2 = \frac{2 \cdot \frac{c}{a}}{-\frac{b}{a}} = \frac{-2c}{b} \]

### Step 5: Form the quadratic equation with roots \( x_1 \) and \( x_2 \)
The quadratic equation with roots \( x_1 \) and \( x_2 \) can be expressed as:
\[ x^2 - (x_1 + x_2)x + (x_1 \cdot x_2) = 0 \]

### Step 6: Calculate \( x_1 + x_2 \) and \( x_1 \cdot x_2 \)
1. **Sum of the roots \( x_1 + x_2 \)**:
\[ x_1 + x_2 = -\frac{b}{2a} + \left(-\frac{2c}{b}\right) = -\frac{b}{2a} - \frac{2c}{b} \]
To combine these fractions, we need a common denominator:
\[ x_1 + x_2 = -\frac{b^2 + 4ac}{2ab} \]

2. **Product of the roots \( x_1 \cdot x_2 \)**:
\[ x_1 \cdot x_2 = \left(-\frac{b}{2a}\right) \cdot \left(-\frac{2c}{b}\right) = \frac{bc}{2a} \]

### Step 7: Substitute into the quadratic equation
Substituting \( x_1 + x_2 \) and \( x_1 \cdot x_2 \) into the quadratic equation:
\[ x^2 - \left(-\frac{b^2 + 4ac}{2ab}\right)x + \frac{bc}{2a} = 0 \]
Multiplying through by \( 2ab \) to eliminate the fractions:
\[ 2abx^2 + (b^2 + 4ac)x - bc = 0 \]

### Final Quadratic Equation
Thus, the quadratic equation whose roots are \( x_1 \) and \( x_2 \) is:
\[ 2abx^2 + (b^2 + 4ac)x - bc = 0 \]