Question

Find the values of k for each of the following quadratic equations, so that they have two equal roots.
2x2 + kx + 3 = 0​

Answer 1

i)

If 2x

2

+kx+3=0 has equal roots, then b

2

−4ac=0.

Here, a=2,b=k,c=3

x

2

+kx+3=0

b

2

−4ac=0

k

2

−4(2)(3)=0

k

2

=24

k=±

24

=±2

6

ii)

If kx

2

−2kx+6=0 has equal roots, then b

2

−4ac=0.

Here, a=k,b=−2k,c=6

kx(x−2)+6=0

kx

2

−2kx+6=0

4k

2

−4(k)(6)=0

4k

2

−24k=0

4k(k−6)=0

For k=0, equation is not quadratic.

k=6

Answer 2

Answer:

(i) 2x2 + kx + 3 = 0

Here, a = 2, b = k, c = 3

b2 – 4ac = (k)2 – 4(2) (3)

= k2 – 24

It has equal roots.

Hence b2 – 4ac = 0.

K^2-24=0

K^2 =24

k=±square root 24