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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.
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For real and equal roots, $D = b^2 - 4ac = 0$
$36 (k + 1)^2 - 4 (k + 1)\times 3 (k + 9) = 0$
$\Rightarrow k^2 - 2k - 3 = 0$
$\Rightarrow (k - 3) (k + 1) = 0$
$k \neq - 1$ So, $k = 3$
$36 (k + 1)^2 - 4 (k + 1)\times 3 (k + 9) = 0$
$\Rightarrow k^2 - 2k - 3 = 0$
$\Rightarrow (k - 3) (k + 1) = 0$
$k \neq - 1$ So, $k = 3$