For what value of k, are the roots of the quadratic equation kx (x-2) +6 = 0equal ?
For Equal roots D = 0kx(x-2) + 6 = 0kx2 -2kx + 6 = 0comparing with ax2+ bx + c = 0, we get\xa0a = k, b = -2k , c = 6D = b2 – 4acPut value of a b c(-2k)2 – 4*6*k = 04k2 – 24k = 04k(k-6) =0\xa04k = 0 or k-6=0k = 6
The given equation is `kx^(2)-2kx+6=0.`
This is of the form `ax^(2)+bx+c=0,` where a=k, b=-2k and c=6.
`:.” “D=(b^(2)-4ac)=(4k^(2)-4xxkxx6)=(4k^(2)-24k).`
For equal roots, we must have
`D=0implies4k^(2)-24k=0implies4k(k-6)=0impliesk=0″ or “k=6.`
Now, k=0, we get 6=0, which is absurd.
`:.” “kne0` andm hance k=6.