If the roots of the equation x2+a2=8x+6a are real, then a belongs to

Answer 1

I will assume you meant
x^2 + a^2 = 8x + 6a.

then
x^2 - 8x + a2 - 6a - 0

to have real roots, the discriminant ≥ 0

64 - 4(1)(a^2 - 6a) ≥ 0
64 - 4a^2 + 24a ≥ 0
a^2 - 6a - 16 < 0
(a-8)(a+2) ≤ 0

-2 ≤ a ≤ 8, a is a real number.

Answer 2

The given equation can be written as
x^2-8x+a-6a=0
Since the roots of the above equation are real
B^2-4ac>0
64-4(a^2-6a)>0
a^2-6a-16<0
(a+2)(a-8)<0
(-2,8)