Question

find the value of K for which Q. E has real and equal root Kx (x-2√5)+10=0​

Answer 1

this may help you.....

Answer 2

Answer:

k = 2

Step-by-step explanation:

kx (x-2√5)+10 = 0

kx^2 - 2√5kx + 10 = 0

If roots are real and equal then b^2 - 4ac = 0

b = (-2√5k)

a = k

c = 10

(-2√5k)^2 - 4×k×10 = 0

4×5×k^2 - 40k = 0

20k^2 - 40k = 0

20k(k - 2) = 0

k - 2 = 0/20k

k -2 = 0

k = 2