How do we show that the expression x^2 + 2ax + b is no less than b^2 - a^2?

Question

If we take this has/have real root(s) we can tell that 4a^2 - 4c>= 0.
But how do we prove the given?

Answer 1

There must be some error here.
Suppose x = 5, a = -1, b = -10.

Then:
x ² + 2ax + b
= 5 ² + 2(-1)(5) + (-10)
= 25 - 10 - 10
= 5

And:
b ² -a ²
= (-10) ² - (-1) ²
= 100 - 1
= 99

5 < 99

Thus x ² + 2ax + b can be less than b ² -a ².