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

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

Here, $a = 1$, $b = -2(p+1)$, $c = p^2$

Condition for real roots: Discriminant $D \geq 0$

$$D = b^2 - 4ac = [-2(p+1)]^2 - 4(1)(p^2)$$

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

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

$$= 4p^2 + 8p + 4 - 4p^2$$

$$= 8p + 4$$

For real roots: $8p + 4 \geq 0$

$$\Rightarrow p \geq -\frac{1}{2}$$

Smallest value of $p$ = $-\dfrac{1}{2}$

Finding roots when $p = -\dfrac{1}{2}$:

The equation becomes:

$$x^2 - 2\left(-\frac{1}{2}+1\right)x + \left(-\frac{1}{2}\right)^2 = 0$$

$$x^2 - 2\left(\frac{1}{2}\right)x + \frac{1}{4} = 0$$

$$x^2 - x + \frac{1}{4} = 0$$

$$\left(x - \frac{1}{2}\right)^2 = 0$$

$$\therefore x = \frac{1}{2}, \frac{1}{2}$$

The roots of the equation are $\dfrac{1}{2}$ and $\dfrac{1}{2}$ (equal roots).

Source: Chapter 4, Section 4.4 – Nature of Roots

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• Examiners want you to clearly write the discriminant condition ($D \geq 0$ for real roots), expand it step by step, and solve the inequality — each step carries marks.
• The word "smallest value" signals you should set $D = 0$ (boundary case), giving equal/coincident roots.
• After finding $p = -\frac{1}{2}$, substitute back and solve — don't forget to show the factored form or simplification.
• Writing the final roots clearly as two equal values of $\frac{1}{2}$ is essential for full marks.