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 start with the two quadratic equations given:

1. \( 4x^2 - 2x - m = 0 \) (Equation 1)
2. \( 4pq - r \cdot x^2 - 2pr - p \cdot x + r(p - q) = 0 \) (Equation 2)

We know that these equations have a common root, and that Equation 2 has equal roots. Let's denote the common root as \( \alpha \).

### Step 1: Analyze Equation 2 for Equal Roots
For Equation 2 to have equal roots, the discriminant must be zero. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[ D = b^2 - 4ac \]

For Equation 2:
- \( a = 4pq - r \)
- \( b = -2pr - p \)
- \( c = r(p - q) \)

Setting the discriminant to zero:

\[ (-2pr - p)^2 - 4(4pq - r)(r(p - q)) = 0 \]

### Step 2: Substitute the Common Root
Since \( \alpha \) is a common root, it must satisfy both equations. We will first substitute \( \alpha \) into Equation 2.

Assuming \( \alpha = -\frac{1}{2} \) (as suggested in the video transcript), we substitute this value into Equation 2:

\[ 4pq - r \left(-\frac{1}{2}\right)^2 - 2pr \left(-\frac{1}{2}\right) - p \left(-\frac{1}{2}\right) + r(p - q) = 0 \]

Calculating each term:

1. \( 4pq \)
2. \( -r \cdot \frac{1}{4} \)
3. \( +pr \)
4. \( +\frac{p}{2} \)
5. \( +r(p - q) \)

Combining these terms gives us an equation in terms of \( p, q, r \).

### Step 3: Solve for \( m \) Using Equation 1
Now, we substitute \( \alpha = -\frac{1}{2} \) into Equation 1:

\[ 4\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) - m = 0 \]

Calculating each term:

1. \( 4 \cdot \frac{1}{4} = 1 \)
2. \( +1 \) (from \( -2 \cdot -\frac{1}{2} \))
3. Thus, we have \( 1 + 1 - m = 0 \)

This simplifies to:

\[ 2 - m = 0 \]

### Step 4: Solve for \( m \)
Rearranging gives:

\[ m = 2 \]

### Final Answer
The value of \( m \) is \( 2 \).

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