For which values of $p$ does the quadratic equation $x^{2} - (2 p + 3) x + (p^{2} + 3 p + 2) = 0$ have real and distinct roots?

medium MCQ 2 marks Class 10 · Mathematics · Quadratic Equations

Answer

All real values of p

The correct answer is All real values of p — Therefore, all real numbers $p$ satisfy the condition.

Step-by-step solution

The discriminant $D$ of $a x^{2} + b x + c = 0$ is $D = b^{2} - 4 a c$ .
Here $a = 1$ , $b = - (2 p + 3)$ , $c = p^{2} + 3 p + 2$ . Compute $D$ :
$D = (2 p + 3)^{2} - 4 \cdot 1 \cdot (p^{2} + 3 p + 2)$
Expand: $D = 4 p^{2} + 12 p + 9 - 4 p^{2} - 12 p - 8$
Simplify: $D = 1$ .
Since $D = 1 > 0$ for every real $p$ , the equation always has two distinct real roots.
Therefore, all real numbers $p$ satisfy the condition.

Common mistake: Forgetting to square the negative sign of

b

, leading to an incorrect discriminant.

Concept tested: Nature of roots via discriminant