(i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
Solution:
(i) 2x2 + kx + 3 = 0
Comparing the given equation with ax2 + bx + c = 0, we get;
a = 2, b = k and c = 3
As we know, Discriminant = b2 – 4ac
= (k)2 – 4(2) (3)
= k2 – 24
For equal roots, we know,
Discriminant = 0
k2 – 24 = 0
k2 = 24
k = ±√24 = ±2√6
(ii) kx(x – 2) + 6 = 0
or kx2 – 2kx + 6 = 0
Comparing the given equation with ax2 + bx + c = 0, we get;
a = k, b = – 2k and c = 6
We know, Discriminant = b2 – 4ac
= ( – 2k)2 – 4 (k) (6)
= 4k2 – 24k
For equal roots, we know,
b2 – 4ac = 0
4k2 – 24k = 0
4k (k – 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms x2 and x.
Therefore, if this equation has two equal roots, k should be 6 only.