Proof that $f(x)=x^4+5x^2-9=0$ has at least two real roots and another question? So I had my exam for real analysis. I'd just like to make sure if one of my answers was right or not.
I haven't seen any question like this around the internet and I've never had to do this problem in class before, so I thought of the answer on the spot. I wrote nearly a page for this answer, but here's the quick gist of it.
Proof attempt:
We know that $f(x)$ is continuous for all real numbers since it's a sum of continuous functions and a constant. So, let $x=0$. Then $f(0)=-9$. Let $x=2$. Then $f(2)=27$. So, by the intermediate value theorem, there $\exists x_0$ such that $f(x_0)=0$. So, this proves that there is at least one real root.
Since the function's variables are all even powered, it follows that $x^4$=$(-x)^4$ and $5x^2=5(-x)^2$. So, $x^4+5x^2-9=(-x)^4+5(-x)^2-9$. Hence, let $x_n=-x_0$. Then, this would also be a root.
So, there $\exists x_0$ and $x_n$, which are two real roots. This proves there are at least two real roots.
Should I have explained more about the even powered variables? Some of my friends did it by contradictions.
(BONUS) There was also another question that said something around lines of, $f(x)$ is a continuous function where $f(x_0) \gt 0$. Prove there $\exists x$ such that $x \in (x_0-\epsilon, x_0+ \epsilon)$ for some $\epsilon \gt 0$.
For this, I went the with the $\epsilon - \delta$ definition where $|x-x_0| \lt \delta$ and showed that since $|x| \lt \delta -|x_0|$, it implies $x_0-\delta \lt x \lt x_0+ \delta $. If you don't get what I'm saying, then that's fine. i feel like I did that wrong...
 A: Let's walk through the proof, shall we?


We know that $f(x)$ is continuous for all real numbers since it's a sum of continuous functions and a constant.

Small detail: a constant is also a continuous function, but what you said isn't wrong.

So, let $x=0$. Then $f(0)=-9$. Let $x=2$. Then $f(2)=27$. So, by the intermediate value theorem, there $\exists x_0$ such that $f(x_0)=0$. So, this proves that there is at least one real root.

Perfect.

Since the function's variables are all even powered, it follows that $x^4$=$(-x)^4$ and $5x^2=5(-x)^2$. So, $x^4+5x^2-9=(-x)^4+5(-x)^2-9$. Hence, let $x_n=-x_0$. Then, this would also be a root.

Why you would call one root $x_0$ and the other $x_n$ is a mystery to me, but nevertheless, it's not incorrect. The function being even indeed means the roots are mirrored over $0$, so if $x_0$ is a root, so is $-x_0$.

So, there $\exists x_0$ and $x_n$, which are two real roots. This proves there are at least two real roots.

Again, a small detail; we could have $x_0=x_n$, but this is easily disproved by showing $x_0\neq 0$.

Should I have explained more about the even powered variables? Some of my friends did it by contradictions.

No, this is perfectly fine. It of course depends on the context, but your proof is very clear. You should be okay.

As far as the bonus goes:


$f(x)$ is a continuous function where $f(x_0) \gt 0$. Prove there $\exists x$ such that $x \in (x_0-\epsilon, x_0+ \epsilon)$ for some $\epsilon \gt 0$.

I don't think you meant that; sure, take $x=x_0$, and $x \in (x_0-\epsilon, x_0+ \epsilon)$.
A: Your proof is fine.
Another way of looking at this is as the composition of two functions - take $y(x)=x^2$ then $f(x)=y^2+5y-9=g(y)$
Now observe that if $g(y)=0$ has a positive root $y_0$, then $y=x^2$ has two real roots $x=\pm \sqrt {y_0}$ and these are roots of $f(x)=0$.
Now you are simply in the business of showing that a particular quadratic has a positive real root. Either of the methods suggested - solving the quadratic, or using the intermediate value theorem as you have done will be sufficient. I note that you don't need to use the full algebraic solution of the quartic for that method - just to identify a positive root of the quadratic.
My instinct was to solve the quadratic. For more complex cases of this kind note (a) using this composition trick definitely might save you arithmetic even if you are using IVT (if $x=0, \pm 1$ it is easy, but for other values of the variable you have lower powers to compute) and (b) that you wouldn't be going for the algebraic solution of a higher degree polynomial unless you could spot an easy root and factorise it. The Intermediate Value Theorem works all the time, but you have to find the right values to try.
[Rather too long for a comment.]
A: Your proof is valid, but there are commonly accepted facts that you don't seem to be aware of. First, all polynomials are continuous (given basic assumptions such as we're dealing with real numbers, etc.), so your instructor would probably have accepted you just saying "This is continuous because it's a polynomial". Second, the terms "odd" and "even" are commonly used to describe functions with the property $f(-x)=-f(x)$ and $f(-x)=f(x)$, respectively, and it's known that polynomials composed entirely of even powers are even in this sense (and odd powers give an odd function). So your instructor probably would have accepted "We have $x_0$ such that $f(x_0)=0$, $f$ is even, thus $f(-x_0)=0$". 
You do have a slight omission, however, in that you did not take note of the possibility that $x_0$ and $-x_0$ are the same number. This can be dismissed based on the argument that $x_0>0$, but it would have been good to explicitly make that argument.
As for your second question, it doesn't make any sense. That, given any $x_0$, there exist $x,\epsilon$ such that $x$ is within $\epsilon$ of $x_0$ is trivially true. My guess is that the problem asked to prove $\exists \delta: \forall x:  (x \in (x_0-\delta,x_0+\delta))\rightarrow (f(x)>0)$. This can be proven by taking the epsilon-delta definition of continuity, taking $\epsilon$ to be $f(x_0)$. There is some $\delta$ such that for all $x$ within $\delta$ of $x_0$, $f(x)$ is within $\epsilon$ of $f(x_0)$. Then since $\epsilon=f(x_0)$, we have that $f(x)$ is within $f(x_0)$ of $f(x_0)$, and it easily follows that $f(x)>0$.
