Find the value of $k$ for which the quadratic equation $(k - 2) x^{2} + 2 (2 k - 3) x + (5 k - 6) = 0$ has real and equal roots.

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Solution

$\Rightarrow$ The given quadratic equation is $(k + 1) x^{2} - (5 k - 3) + 9 = 0$ , comparing it with $a x^{2} + b x + c = 0$
We get, $a = k - 2, b = 2 (2 k - 3), c = 5 k - 6$
It is given that roots are real and equal.
$∴$ $b^{2} - 4 a c = 0$
$\Rightarrow$ $[2 (2 k - 3)]^{2} - 4 (k - 2) (5 k - 6) = 0$
$\Rightarrow$ $16 k^{2} - 48 k + 36 - 20 k^{2} + 64 k - 48 = 0$
$\Rightarrow$ $4 k^{2} - 16 k + 12 = 0$
$\Rightarrow$ $k^{2} - 4 k + 3 = 0$
$\Rightarrow$ $(k - 3) (k - 1) = 0$
$∴$ $k = 3$ or $k = 1$

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