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

For the equation $(c+1)x^2 - 6(c+1)x + 3(c+9) = 0$ to have real and equal roots, the discriminant must be zero.

Here, $a = (c+1)$, $b = -6(c+1)$, $c = 3(c+9)$.

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

$$[-6(c+1)]^2 - 4 \cdot (c+1) \cdot 3(c+9) = 0$$

$$36(c+1)^2 - 12(c+1)(c+9) = 0$$

$$12(c+1)[3(c+1) - (c+9)] = 0$$

$$12(c+1)[3c + 3 - c - 9] = 0$$

$$12(c+1)(2c - 6) = 0$$

$$24(c+1)(c-3) = 0$$

So, $c+1 = 0$ or $c - 3 = 0$, giving $c = -1$ or $c = 3$.

Since $c \neq -1$ (given), $c = 3$.

Source: Nature of Roots, Chapter 4, Section 4.4

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• The key condition for equal roots is discriminant = 0, i.e., $b^2 - 4ac = 0$.
• Identify $a$, $b$, $c$ carefully from the given equation before substituting.
• After factorising, reject $c = -1$ because the problem states $c \neq -1$ (if $c = -1$, the equation stops being quadratic).
• Examiners award marks stepwise: setting up discriminant, simplifying, factorising, and stating the valid answer — show every step clearly.