Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x - 2) + 6 = 0
(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b2 - 4ac
= (k)2 - 4(2) (3)
= k2 - 24
For equal roots,
Discriminant = 0
k2 - 24 = 0
k2 = 24
k = ±√24 = ±2√6
(ii) kx(x - 2) + 6 = 0
or kx2 - 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = k, b = - 2k and c = 6
Discriminant = b2 - 4ac
= ( - 2k)2 - 4 (k) (6)
= 4k2 - 24k
For equal roots,
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.
