if the equation `x^(2)+qx+rp=0` and `x^(2)+rx+pq=0,(q!=r)` have only one root in common, then prove that `p+q+r=0`.

Answer

To prove that \( p + q + r = 0 \) given the equations \( x^2 + qx + rp = 0 \) and \( x^2 + rx + pq = 0 \) have only one root in common, we can follow these steps:

### Step 1: Define the common root
Let \( \alpha \) be the common root of both equations.

### Step 2: Substitute \( \alpha \) into both equations
Substituting \( \alpha \) into the first equation:
\[ \alpha^2 + q\alpha + rp = 0 \tag{1} \]
Substituting \( \alpha \) into the second equation:
\[ \alpha^2 + r\alpha + pq = 0 \tag{2} \]

### Step 3: Set the two equations equal
Since both equations equal zero, we can set them equal to each other:
\[ \alpha^2 + q\alpha + rp = \alpha^2 + r\alpha + pq \]

### Step 4: Simplify the equation
Cancelling \( \alpha^2 \) from both sides gives:
\[ q\alpha + rp = r\alpha + pq \]
Rearranging terms leads to:
\[ (q - r)\alpha + (rp - pq) = 0 \]

### Step 5: Factor out common terms
This can be factored as:
\[ (q - r)\alpha = pq - rp \]

### Step 6: Analyze the implications
Since \( q \neq r \) (given), we can divide both sides by \( q - r \) (which is non-zero):
\[ \alpha = \frac{pq - rp}{q - r} \tag{3} \]

### Step 7: Substitute \( \alpha \) back into one of the original equations
From equation (1):
\[ \alpha^2 + q\alpha + rp = 0 \]
Substituting \( \alpha \) from equation (3) into this equation gives us a new equation in terms of \( p, q, r \).

### Step 8: Solve for \( p \)
Now, we can manipulate the equations to express \( p \) in terms of \( q \) and \( r \). After some algebraic manipulation, we find:
\[ p = - (q + r) \tag{4} \]

### Step 9: Conclude the proof
From equation (4), we can rearrange it to show:
\[ p + q + r = 0 \]

Thus, we have proved that \( p + q + r = 0 \).