For what values of 'k' does the equation $(k - 2) x^{2} + 2 (2 k - 3) x + (5 k - 6) = 0$ have equal roots ? Find the roots of the equations corresponding to those value of k ?

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Solution

$D = 0$
$b^{2} - 4 a c = 0$
${[2 (2 k - 3)]}^{2} - 4 (k - 2) (5 k - 6) = 0$
$\Rightarrow 4 (2 k - 3)^{2} - 4 [5 k^{2} - 6 k - 10 k + 12] = 0$
$\Rightarrow 4 (4 k^{2} - 12 k + 9) - 4 [5 k^{2} - 16 k + 12] = 0$
$\Rightarrow 16 k^{2} - 48 k + 36 - 20 k^{2} + 64 k - 48 = 0$
$\Rightarrow - 4 k^{2} + 16 k - 12 = 0$
$\Rightarrow k^{2} - 4 k + 3 = 0$
$\Rightarrow (k - 3) (k - 1)$
$\Rightarrow k = 3, 1$
When $k = 1$ , the equation will be:
$- x^{2} - 2 x - 1 = 0$
$x^{2} + 2 x + 1 = 0$
$(x + 1)^{2} = 0$
$x = - 1, - 1$

$k = 1$ , corresponding roots are -1, -1
$k = 3$ , corresponding roots are -3, -3

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