(i) 2x^{2} + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

**Solution:**

**(i) 2 x^{2} + kx + 3 = 0**

Comparing the given equation with *ax*^{2} + *bx *+ *c* = 0, we get;

*a* = 2, *b* = k and *c* = 3

As we know, Discriminant = *b*^{2} – 4*ac*

= (*k*)^{2} – 4(2) (3)

= *k*^{2} – 24

For equal roots, we know,

Discriminant = 0

*k*^{2} – 24 = 0

*k*^{2} = 24

k = ±√24 = ±2√6

**(ii) kx(x – 2) + 6 = 0**

or *kx*^{2} – 2*kx* + 6 = 0

Comparing the given equation with *ax*^{2} + *bx *+ *c* = 0, we get;

*a* = *k*, *b* = – 2*k* and *c* = 6

We know, Discriminant = *b*^{2} – 4*ac*

= ( – 2*k*)^{2} – 4 (*k*) (6)

= 4*k ^{2}* – 24

*k*

For equal roots, we know,

*b*^{2} – 4*ac* = 0

4*k*^{2} – 24*k* = 0

4*k* (*k* – 6) = 0

Either 4*k* = 0 or *k* = 6 = 0

*k* = 0 or *k* = 6

However, if *k* = 0, then the equation will not have the terms *x*^{2} and *x*.

Therefore, if this equation has two equal roots, *k* should be 6 only.