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Find the value of 'k' for which the quadratic equation $(k + 1) x^2 - 2 (3k + 1) x + (8k + 1) = 0$ has real and equal roots.
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For real and equal roots.
$[-2(3k + 1)]^2 - 4(k + 1)(8k + 1) = 0$
$\Rightarrow k^2 - 3k = 0$
$\therefore k = 0, k = 3$
$[-2(3k + 1)]^2 - 4(k + 1)(8k + 1) = 0$
$\Rightarrow k^2 - 3k = 0$
$\therefore k = 0, k = 3$