Question

For what value of k are the roots of the quadratic equation $kx(x-2\sqrt{5})+10=0$ real and equal?

Answer 1

D= b²-4ac is called discriminant.

The nature of roots depend upon the value of the discriminant D. Since D can be zero, positive or negative.

When D=0, Roots are real and equal.

kx(x-2√5)+10=0

kx²-2√5kx+10=0

Here, a= k, b= -2√5k, c= 10

D= b²-4ac=0

(-2√5k)² - 4×k×10=0

4×5k² -40k =0

20k²-40k=0

20k(k-2)=0

k(k-2)=0

k≠0

k-2=0

k=2

Hence, the value of k is 2.