prove quadratic polynomial has no real roots

Question

The problem asks me to prove that a polynomial $f(x)=x^2+ax+b$ has no real roots for some $a,b \in \Bbb{R}$

I started by assuming that $f(x)=x^2+ax+b$ has real roots and therefore the determinant $a^2-4b \geq 0$ and that $a \geq \pm 2 \sqrt{b}$ which imples that $b \geq 0$ which contradicts the fact that $b \in \Bbb{R}$ is this a correct way of doing this proof?

Answer 1

Guide:

You just have to show that the statement is true for some $a,b$.

Why dont' we set $a=0$, now can you choose a $b$ such that $x^2+b$ has no real root?

Remark about your attempt:

From $$a^2 \geq 4b,$$ when you take square root, it seems that you have assumed that $a \geq 0$ and $b \geq 0$.

Answer 2

Consider an example where a=2*sqrt(b). Then we have something that we can factor into (x+sqrt(b))^2, and letting this equal zero, we get a real root for x, -sqrt(b). I may have made a mistake in my understanding of the problem, but it doesn't seem to be a correctly worded problem.