Find the value of '$k$' for which the quadratic equation $(k + 1)x^2 - 6(k + 1)x + 3(k + 9) = 0$, $k \neq -1$ has real and equal roots.
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
For a quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots, the discriminant must be zero:
$$D = b^2 - 4ac = 0$$
Given equation: $(k+1)x^2 - 6(k+1)x + 3(k+9) = 0$
Here, $a = (k+1)$, $b = -6(k+1)$, $c = 3(k+9)$
Setting $D = 0$:
$$[-6(k+1)]^2 - 4(k+1) \cdot 3(k+9) = 0$$
$$36(k+1)^2 - 12(k+1)(k+9) = 0$$
$$12(k+1)[3(k+1) - (k+9)] = 0$$
$$12(k+1)[3k + 3 - k - 9] = 0$$
$$12(k+1)(2k - 6) = 0$$
$$24(k+1)(k-3) = 0$$
So, $k + 1 = 0$ or $k - 3 = 0$, giving $k = -1$ or $k = 3$.
Since $k \neq -1$ (given), therefore $k = 3$.
Source: Chapter 4, Section 4.4 – Nature of Roots
---
Explanation
- 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, always reject the excluded value ($k = -1$ here) and state the valid answer explicitly.
- Examiners award marks at each step: identifying $a, b, c$ (1M), setting up $D=0$ (1M), simplification (2M), final answer with rejection of $k=-1$ (1M).