Question

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

Answer 1

(k-2)x²+2(2k-3)x+(5k-6)

=>If the equation has equal and real roots than the value of it's discriminant will equal to 0.

D=b²-4ac
0={2(2k-3)}²-4(k-2)(5k-6)

0={4k-6}²-(4k+8)(5k-6)

0=16k²+36-48k-20k²+24k+40k-48
0=-4k²-12+16k
4k²+12-16k=0

k²-4k+3=0

k²-3k-k+3=0

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

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

(k-1)=0
k=1

(k-3)=0
k=3

For both the value of k the quadratic equation has real and equal roots.


@Altaf