Find the value of $p$ for which the quadratic equation $(p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0$ has equal roots. Hence find the roots.
Step-by-Step Solution
Key Concept: $D = 36(p+1)^2 - 12(p+1)(p+9) = 0 \Rightarrow 12(p+1)[3(p+1) - (p+9)] = 0 \Rightarrow 12(p+1)(2p - 6) = 0 \Rightarrow p = 3$ (since $p <br>eq -1$). Roots: $4x^2 - 24x + 36 = 0 \Rightarrow (2x - 6)^2 = 0 \Rightarrow x = 3$.
$36(p+1)^2 - 12(p+1)(p+9) = 0 \Rightarrow 12(p+1)(2p - 6) = 0 \Rightarrow p = 3$. [1.5 Marks]
For $p = 3$: $4x^2 - 24x + 36 = 0 \Rightarrow x^2 - 6x + 9 = 0 \Rightarrow (x - 3)^2 = 0 \Rightarrow x = 3, 3$. [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Solving $p = 3$: 1.5 Marks
Finding equal roots $x = 3, 3$: 1.5 Marks
Correct Answer: