The roots of the equation `(x-p) (x-q)=r^(2)` where p,q , r are real , are

Options

A. always complex
B. always real
C. always purely imaginary
D. None of these

Explanation

To find the roots of the equation \((x - p)(x - q) = r^2\), we will first rewrite it in standard quadratic form and then analyze the nature of the roots using the discriminant.

### Step-by-Step Solution:

1. **Expand the Equation**:
Begin by expanding the left side of the equation:
\[ (x - p)(x - q) = x^2 - (p + q)x + pq \]
Therefore, the equation becomes:
\[ x^2 - (p + q)x + pq = r^2 \]

2. **Rearrange to Standard Form**:
Rearranging the equation gives us:
\[ x^2 - (p + q)x + (pq - r^2) = 0 \]
This is now in the standard quadratic form \(ax^2 + bx + c = 0\), where:
- \(a = 1\)
- \(b = -(p + q)\)
- \(c = pq - r^2\)

3. **Calculate the Discriminant**:
The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (-(p + q))^2 - 4(1)(pq - r^2) \]
Simplifying this gives:
\[ D = (p + q)^2 - 4(pq - r^2) \]

4. **Further Simplification**:
Expanding the discriminant:
\[ D = (p^2 + 2pq + q^2) - (4pq - 4r^2) \]
This simplifies to:
\[ D = p^2 + q^2 + 2pq - 4pq + 4r^2 = p^2 + q^2 - 2pq + 4r^2 \]

5. **Final Form of Discriminant**:
The expression can be rearranged as:
\[ D = (p - q)^2 + 4r^2 \]

6. **Determine the Nature of Roots**:
Since both \((p - q)^2\) and \(4r^2\) are always non-negative (as they are squares), we conclude:
\[ D \geq 0 \]
This indicates that the roots of the equation are real.

### Conclusion:
The roots of the equation \((x - p)(x - q) = r^2\) are real.