Question

The roots of the quadratic equation \(x^2 + x – p (p + 1) = 0 \)are :

• A. p, p + 1
• B. – p, p + 1
• C. – p, – (p + 1)
• D. p, – ( p + 1)

Answer 1

The Correct Option is D

Step 1: Understanding the problem:
We are given the quadratic equation:
\[ x^2 + x - p(p + 1) = 0 \] We are asked to find the roots of the quadratic equation in terms of \( p \).

Step 2: Using the quadratic formula:
The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where the roots are given by the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \( x^2 + x - p(p + 1) = 0 \), we have:
- \( a = 1 \) - \( b = 1 \) - \( c = -p(p + 1) \) Substitute these values into the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-p(p + 1))}}{2(1)} \] Simplifying: \[ x = \frac{-1 \pm \sqrt{1 + 4p(p + 1)}}{2} \] \[ x = \frac{-1 \pm \sqrt{1 + 4p^2 + 4p}}{2} \] \[ x = \frac{-1 \pm \sqrt{(2p + 1)^2}}{2} \] Since \( \sqrt{(2p + 1)^2} = |2p + 1| \), we have two cases for the roots:
Case 1: \( 2p + 1 \geq 0 \):
\[ x = \frac{-1 + (2p + 1)}{2} = \frac{2p}{2} = p \] Case 2: \( 2p + 1 < 0 \):
\[ x = \frac{-1 - (2p + 1)}{2} = \frac{-2p - 2}{2} = -(p + 1) \] Thus, the roots of the quadratic equation are \( p \) and \( -(p + 1) \).

Conclusion:
The roots of the quadratic equation \( x^2 + x - p(p + 1) = 0 \) are \( p \) and \( -(p + 1) \).