Mathematics — Standard (041) — CBSE Class 10 board question

Given equation: $(p + 4)x^2 - (p + 1)x + 1 = 0$

Here, $a = (p + 4)$, $b = -(p + 1)$, $c = 1$

Note: For it to be a quadratic equation, $p + 4 \neq 0$, i.e., $p \neq -4$.

For real and equal roots, discriminant $= 0$:

$$b^2 - 4ac = 0$$

$$[-(p+1)]^2 - 4(p+4)(1) = 0$$

$$(p+1)^2 - 4(p+4) = 0$$

$$p^2 + 2p + 1 - 4p - 16 = 0$$

$$p^2 - 2p - 15 = 0$$

$$p^2 - 5p + 3p - 15 = 0$$

$$(p-5)(p+3) = 0$$

$$\therefore p = 5 \quad \text{or} \quad p = -3$$

Finding roots:

Case 1: When $p = 5$, the equation becomes:
$$9x^2 - 6x + 1 = 0 \implies (3x-1)^2 = 0$$
$$x = \frac{1}{3}, \frac{1}{3}$$

Case 2: When $p = -3$, the equation becomes:
$$x^2 - (-2)x + 1 = 0 \implies x^2 + 2x + 1 = 0 \implies (x+1)^2 = 0$$
$$x = -1, -1$$

Source: Chapter 4, Section 4.4 (Nature of Roots)

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• Examiners award marks for: setting up D = 0 correctly (1 mark), expanding and simplifying (1 mark), solving for $p$ (1 mark), finding roots for each value of $p$ (2 marks).
• Don't forget to check $p \neq -4$ (otherwise the equation won't be quadratic). Mentioning this shows conceptual clarity.
• Equal roots mean $x = -\frac{b}{2a}$ (repeated), but verifying by factorisation (as shown) is cleaner and often expected at this level.