find the values of k,for which the quadratic equation has equal roots:kx(x-2)+6(k-2)

Asked by  | 9th Jun, 2008, 04:23: PM

Expert Answer:

kx(x-2)+6(k-2)

= kx2 - 2kx + 6(k-2)

a = k , b = -2k , c = 6(k-2)

If the discriminant b2 - 4ac equals zero, the radical in the quadratic formula becomes zero.

b2 - 4ac = (-2k)2 - 4.k.6(k-2)

= 4k2 - 24k2 + 48k

= -20k2 + 48k = 0

-5k2 + 12k = 0

-k (5k -12) = 0

k = 0, 12/5

Answered by  | 9th Jun, 2008, 05:00: PM

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