Find the value of $k$ for which the quadratic equation $(k + 1)x^2 - 6(k + 1)x + 3(k + 9) = 0$, $k
eq -1$ has real and equal roots.

Question

Find the value of $k$ for which the quadratic equation $(k + 1)x^2 - 6(k + 1)x + 3(k + 9) = 0$, $k 
eq -1$ has real and equal roots.

Accepted Answer

Step 1: For real and equal roots, $D = b^2 - 4ac = 0$: $36(k + 1)^2 - 4(k + 1) 	imes 3(k + 9) = 0$
Step 2: $36(k + 1) - 12(k + 9) = 0 \Rightarrow 36k + 36 - 12k - 108 = 0 \Rightarrow 24k - 72 = 0$
Step 3: $\Rightarrow k = 3$

Question

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