Answer 1

The given quadric equation is `kx^2-2sqrt5x+4=0`, and roots are real and equal

Then find the value of k.

Here, a = k, `b=-2sqrt5`, and c = 4

As we know that D = b² - 4ac

Putting the value of a = k, `b=-2sqrt5`, and c = 4

`=(-2sqrt5)^2-4xxkxx4`

= 20 - 16k

The given equation will have real and equal roots, if D = 0

Thus,

20 - 16k = 0

16k = 20

k = 20/16

k = 5/4

Therefore, the value of k = 5/4.