Quadratic equation!!

x^2+2kx+k(k-1)=0△ = (2k)^2 - 4k(k-1)= 4k^2 - 4k^2 + 4k= 4k4k > 0 for any positive values of k.Hence x^2+2kx+k(k-1)=0 has two distinct roots for any positive values of k
since △ > 0

prove that the quadratic equation x^2+2kx+k(k-1)=0 has two distinct roots for any positive values of k.
b^2-4ac
=4k^2-4k(k-1)
=4k^2-4k^2+4k
=4k

4k>0 for any positive values of k


If the quadratic equation 6x^2-7X-1/3=2k has 2 distinct real roots,
i) Find the range of possible values of k

b^2>4ac7^2>-4(6)(1/3+2k)49>-8-48k48k>-57k>-57/48k>-19/16

ii)find the roots of the equation when k is the negative integer in (i)
k>-19/16
k>-1.1875
k=-1 (k is the negative integer)
the quadratic equation 6x^2-7X-1/3=2k
6x^2-7x-1/3=-2
18x^2-21x-1=-6
18x^2-21x+5=0
x=(21+9)/36 or (21-9)/36
x=30/36 or 12/36
x=5/6 or 1/3

b) Using the result of (ii), solve -3(x+1)^2+7/2(x+1)-5/6=0
6(x+1)^2-7(x+1)+5/3=0
x+1=5/6 or x+1=1/3
x=-1/6 or x=-2/3

Δ = (2k)^2 - 4k(k-1)
= 4k^2 - 4k^2 + 4k
= 4k
Since k can be any positive values,
Δ > 0
Hence the equation has two distinct roots.