Question

find the value of K for which quadratic equation ( k-2)x square +2(2x-3) x + (5k -6)=0 has equal roots

Answer 1

$\large\underline{\sf{Solution-}}$

Given that the quadratic equation

$\rm :\longmapsto\:(k - 2) {x}^{2} + 2(2k - 3)x + (5k - 6) = 0$

has equal roots.

So, Discriminant = 0.

Here,

$\rm :\longmapsto\:a = k - 2$

$\rm :\longmapsto\:b = 2(2k - 3)$

$\rm :\longmapsto\:c = 5k - 6$

So,

$\rm :\longmapsto\: {b}^{2} - 4ac = 0$

$\rm :\longmapsto\:{\bigg(2(2k - 3)\bigg) }^{2} - 4(k - 2)(5k - 6) = 0$

$\rm :\longmapsto\:4( {4k}^{2} + 9 - 12k) - 4( {5k}^{2} - 6k - 10k + 12) = 0$

$\rm :\longmapsto\:{4k}^{2} + 9 - 12k -{5k}^{2} + 6k + 10k - 12 = 0$

$\rm :\longmapsto\: - {k}^{2} + 4k - 3 = 0$

$\rm :\longmapsto\: {k}^{2} - 4k + 3 = 0$

$\rm :\longmapsto\: {k}^{2} - 3k - k+ 3 = 0$

$\rm :\longmapsto\:k(k - 3) - 1(k - 3) = 0$

$\rm :\longmapsto\:(k - 3)(k - 1) = 0$

$\bf\implies \:k = 1 \: \: \: or \: \: \: k = 3$