Find the values of k for which the roots are real and equal of the following equations
$x^{2}$ - 2kx + 7k - 12 = 0

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Solution

$x^{2} - 2 k x + 7 k - 12 = 0$
For equal & real roots, $D \equiv 0$
$D \equiv + 4 k^{2} - (7 k - 12) (1) (4)$
$0 \equiv + 4 K^{2} - 28 k + 48$
$0 = + K^{2} - 7 k + 12$
$K^{2} - 4 k - 3 k + 12$
$0 = (k - 4) (k - 3)$
$\Rightarrow k = 4, 3$

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