find the value of K if the quadratic equation ;(K-2)x^2+2(2 k-3)x+(5 K-6)=0

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Solution

The given equation is
(k - 2)x² + 2(2k - 3)x + (5k - 6)
and the roots are real and equal.
here a = k - 2
b = 4k - 6
c = 5k - 6
we konw that
D= b² - 4ac
putting the value of a,b and c

D = (4k - 6)² - 4×( k - 2) ( 5k - 6)
D = 16k² + 36 - 48k - 20k² + 64k - 48
D = 4k² - 16k + 12
taking common 4
acc. to question
k² - 4k+3=0
factories this eq.
(k-3)(k-1)=0
k=3,1.

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