Question

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

Answer 1

Since the equation has equal roots

So,

Discriminent = 0

=> b^2 - 4ac = 0

=> [2(2k-3)]^2 - 4(k-2)(5k-6) = 0

=> 4(4k^2 +9 - 12k) - 4(5k^2 - 6k - 10k +12) = 0

=> 4[ 4k^2 +9 - 12k - (5k^2 - 16k +12)] = 0

=> 4k^2 + 9 - 12k - 5k^2 +16k - 12 = 0

=> - k^2 +4k - 3 = 0

=> k^2 - 4k +3 = 0

=> k^2 - k - 3k + 3 = 0

=> k(k-1) - 3(x-1) = 0

=> (k-1)(k-3) = 0

k = 1 and 3

Answer 2

b^2 - 4ac = 0
(4k-6)^2 - 4(5k-6)(k-2) = 0
k^2-4k+3 = 0
k^2-3k-k+3 = 0
k(k-3)-1(k-3) = 0
(k-1)(k-3) = 0
k = 1 and k = 3