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.

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