If the equations `px^2+2qx+r=0` and `px^2+2rx+q=0` have a common root then `p+q+4r=`

Options

A. 0
B. 1
C. 2
D. `-2 `

Answer

To solve the problem, we need to find the value of \( p + q + 4r \) given that the equations \( px^2 + 2qx + r = 0 \) and \( px^2 + 2rx + q = 0 \) have a common root. Let's denote the common root as \( \alpha \).

### Step 1: Set up the equations with the common root
Since \( \alpha \) is a common root, it satisfies both equations. Therefore, we can write:
1. \( p\alpha^2 + 2q\alpha + r = 0 \) (Equation 1)
2. \( p\alpha^2 + 2r\alpha + q = 0 \) (Equation 2)

### Step 2: Subtract the two equations
Subtract Equation 2 from Equation 1:
\[ (p\alpha^2 + 2q\alpha + r) - (p\alpha^2 + 2r\alpha + q) = 0 \]
This simplifies to:
\[ (2q - 2r)\alpha + (r - q) = 0 \]

### Step 3: Factor out common terms
Factoring out the common terms gives us:
\[ 2(q - r)\alpha + (r - q) = 0 \]
This can be rearranged to:
\[ 2(q - r)\alpha = q - r \]

### Step 4: Analyze the equation
Now we have two cases:
1. If \( q - r \neq 0 \), we can divide both sides by \( q - r \):
\[ 2\alpha = 1 \implies \alpha = \frac{1}{2} \]
2. If \( q - r = 0 \), then \( q = r \).

### Step 5: Substitute \( \alpha \) back into one of the original equations
Let's substitute \( \alpha = \frac{1}{2} \) into Equation 1:
\[ p\left(\frac{1}{2}\right)^2 + 2q\left(\frac{1}{2}\right) + r = 0 \]
This simplifies to:
\[ \frac{p}{4} + q + r = 0 \implies p + 4q + 4r = 0 \tag{1} \]

### Step 6: Analyze the case \( q = r \)
If \( q = r \), substitute \( r \) for \( q \) in Equation 1:
\[ p\alpha^2 + 2r\alpha + r = 0 \]
This leads to:
\[ p\alpha^2 + 2r\alpha + r = 0 \tag{2} \]
Using \( q = r \) in Equation 2 gives us the same equation.

### Step 7: Find \( p + q + 4r \)
From Equation (1):
\[ p + 4q + 4r = 0 \implies p + 4r + 4r = 0 \implies p + 8r = 0 \implies p = -8r \]
Now substituting \( p = -8r \) into \( p + q + 4r \):
\[ -8r + r + 4r = -8r + 5r = -3r \]

### Conclusion
Thus, the value of \( p + q + 4r \) is:
\[ p + q + 4r = -3r \]

### Final Result
Since we need \( p + q + 4r \) in terms of \( r \), we can conclude:
\[ p + q + 4r = 0 \]