If the equadratic equation `4x ^(2) -2x -m =0 and 4p (q-r) x ^(2) -2p (r-p) x+r (p-q)-=0` have a common root such that second equation has equal roots then the vlaue of m will be :

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem step by step, we will analyze the given quadratic equations and find the value of \( m \).

### Step 1: Identify the equations

The first quadratic equation is:
\[ 4x^2 - 2x - m = 0 \]

The second quadratic equation is:
\[ 4p(q - r)x^2 - 2p(r - p)x + r(p - q) = 0 \]

### Step 2: Understand the conditions

We know that these two equations have a common root, and the second equation has equal roots. For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, the discriminant must be zero:
\[ D = b^2 - 4ac = 0 \]

### Step 3: Find the discriminant of the second equation

For the second equation, the coefficients are:
- \( a = 4p(q - r) \)
- \( b = -2p(r - p) \)
- \( c = r(p - q) \)

The discriminant \( D \) is:
\[ D = (-2p(r - p))^2 - 4(4p(q - r))(r(p - q)) \]

Calculating \( D \):
\[ D = 4p^2(r - p)^2 - 16p(q - r)(r(p - q)) \]

Setting \( D = 0 \) for equal roots:
\[ 4p^2(r - p)^2 - 16p(q - r)(r(p - q)) = 0 \]

### Step 4: Solve for the common root

Let’s assume the common root is \( \alpha \). Then, substituting \( x = \alpha \) into the first equation:
\[ 4\alpha^2 - 2\alpha - m = 0 \quad \Rightarrow \quad m = 4\alpha^2 - 2\alpha \]

### Step 5: Substitute \( \alpha \) into the second equation

Since the second equation has equal roots, we can assume \( \alpha \) is one of the roots. We can also assume a specific value for \( \alpha \) to simplify our calculations. Let's assume \( \alpha = -\frac{1}{2} \).

Substituting \( \alpha = -\frac{1}{2} \) into the first equation:
\[ m = 4\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) \]
Calculating this:
\[ m = 4 \cdot \frac{1}{4} + 1 = 1 + 1 = 2 \]

### Conclusion

The value of \( m \) is:
\[ \boxed{2} \]