Question

find positive value of k for which x²-8x+k will have real roots​

Answer 1

Solution:

Compare x²-8x+k = 0 with

ax²+bx+c=0, we get

a = 1 , b = -8 , c = k

Discreminant (D) ≥ 0

[ given , roots are real ]

=> b²-4ac ≥ 0

=> (-8)²-4×1×k≥0

=> 64 - 4k ≥ 0

=> -4k ≥ -64

=> 4k ≤ 64

=> k ≤ 64/4

=> k≤ 16

Therefore,

k≤16