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

We know that if an equation ax² + bx + c = 0 has two equal roots,

its discriminant

(b² − 4ac) will be 0.

(i) 2x² + kx + 3 = 0

Comparing equation with ax² + bx + c = 0, we obtain,

a = 2, b = k, c = 3

Discriminate = b² − 4ac = (k)²− 4(2) (3) = k² − 24

For equal roots,

Discriminant = 0

k² − 24 = 0

k² = 24

=

(ii) kx (x − 2) + 6 = 0

or kx²− 2kx + 6 = 0

Comparing this equation with ax² + bx + c = 0, we obtain,

a = k, b = −2k, 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

= 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.