If p,q and r are rational numbers, then the roots of the equation `x^(2) - 2px + p^(2) + 2 qr - r^(2) = 0` are

Options

A. complex
B. pure imaginary
C. irrational
D. rational

Explanation

To solve the equation \( x^2 - 2px + (p^2 + 2qr - r^2) = 0 \) and determine the nature of its roots, we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be compared with the standard form \( ax^2 + bx + c = 0 \):
- Here, \( a = 1 \)
- \( b = -2p \)
- \( c = p^2 + 2qr - r^2 \)

### Step 2: Calculate the discriminant
The discriminant \( D \) of a quadratic equation is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-2p)^2 - 4(1)(p^2 + 2qr - r^2) \]
Calculating \( D \):
\[ D = 4p^2 - 4(p^2 + 2qr - r^2) \]
\[ D = 4p^2 - 4p^2 - 8qr + 4r^2 \]
\[ D = 4r^2 - 8qr \]
\[ D = 4(r^2 - 2qr) \]

### Step 3: Factor the discriminant
We can factor the expression:
\[ D = 4(r(q - r)) \]

### Step 4: Analyze the discriminant
Since \( p, q, r \) are rational numbers, \( r(q - r) \) is also a rational number. The expression \( r(q - r) \) will be:
- Greater than 0 if \( r \) and \( (q - r) \) have the same sign (both positive or both negative).
- Equal to 0 if \( r = 0 \) or \( q = r \).
- Less than 0 if \( r \) and \( (q - r) \) have opposite signs.

### Step 5: Determine the nature of the roots
- If \( D > 0 \), the roots are real and distinct.
- If \( D = 0 \), the roots are real and equal.
- If \( D < 0 \), the roots are complex.

Since \( D = 4(r(q - r)) \):
- If \( r(q - r) \geq 0 \), the roots are real.
- If \( r(q - r) < 0 \), the roots are complex.

### Conclusion
Given that \( p, q, r \) are rational numbers, if the roots are real, they will also be rational. Therefore, the roots of the equation \( x^2 - 2px + (p^2 + 2qr - r^2) = 0 \) are rational.

### Final Answer
The roots of the equation are rational.