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

Given equation: $px(x - 2) + 6 = 0$

Expanding: $px^2 - 2px + 6 = 0$

Here, $a = p$, $b = -2p$, $c = 6$

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

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

$$(-2p)^2 - 4(p)(6) = 0$$

$$4p^2 - 24p = 0$$

$$4p(p - 6) = 0$$

So, $p = 0$ or $p = 6$.

Since $p = 0$ makes the equation non-quadratic, we reject it.

∴ $p = 6$

Source: Chapter 4, Section 4.4 — Nature of Roots

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• Key condition: Equal real roots ⟹ $b^2 - 4ac = 0$. Examiners expect you to state this explicitly.
• Rejecting p = 0: This step is essential for full marks — if $p = 0$, the equation becomes $6 = 0$, which is not a quadratic. Don't skip this.
• Common mistake: Forgetting to expand the equation first before identifying $a$, $b$, $c$.