Question

Find the value of k for which quadratic equation (k-2)x2 + 2(2k-3) + (5k-6)=0 has equal roots

Answer 1

The given equation is (k-2)x^2+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= b^2-4ac 
putting the value of a,b and c 
D=(4k-6)^2-4×(k-2)(5k-6) 
D=16k^2+36-48k-20k^2+64k-48 
D=4k^2-16k+12 
taking common 4 
acc. to question 
k^2-4k+3=0 
factories this eq. 
(k-3)(k-1)=0 
k=3,1.
I hope it help you

Answer 2

Hello dear,

The given equation is (k-2)x²+2(2k-3)x+(5k-6) and roots are real and equal.
=>a=k-2
=>b=4k-6
=>c=5k-6
=>D= b^2-4ac
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
take common 4
acc. to question
k²-4k+3=0
factories this eq.
(k-3)(k-1)=0
k=3,1