find the value of K for which Q. E has real and equal root Kx (x-2√5)+10=0
Answers
Answered by
0
this may help you.....
Attachments:
Answered by
0
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
Similar questions