Question

If the equation ax^2+bx+c=0 does not have 2 distinct real roots and a+c>b,

then prove that f(x)>=0,for all x belongs to real number.

Answer 1

f(x) = ax² + bx + c?
I'll assume so.
I'll also assume a ≠ 0

Since f(x) has 0 or 1 distinct real roots, then f(x) ≥ 0 for all x iff a > 0

Since f(x) doesn't have 2 distinct real root, discriminant must be ≤ 0.
b² − 4ac ≤ 0
0 ≤ b² ≤ 4ac


If c = 0, then b = 0, and since a+c > b ---> a+0 > 0 ----> a > 0
So f(x) ≥ 0 for all real x


If c ≠ 0, since 4ac ≥ 0, then either a and c are both positive, or a and c are both negative.


If a and c are both positive, then f(x) ≥ 0 for all real x


So the only case that remains is if a and c are both negative.

If a = c, then
0 ≤ b² ≤ 4ac ---> b² ≤ 4a²
b < a+c < 0 ---> b < 2a < 0 ---> b² > 4a²
But b² cannot both be ≤ 4a² and > 4a²

If a ≠ c, then
a−c ≠ 0
(a−c)² > 0
a² − 2ac + c² > 0
a² − 2ac + c² + 4ac > 4ac
a² + 2ac + c² > 4ac
(a+c)² > 4ac
b < a+c < 0 ---> b² > (a+c)² > 4ac
But this contradicts: b² ≤ 4ac

Therefore, a cannot be < 0

Since a > 0 and f(x) has at most 1 real distinct root, then f(x) ≥ 0 for all real x