Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2 (using proof by contradiction). 
The question is "Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2."

I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a well written solution. I would like to know if I solved it right and if I did, I would like some advice on how I can improve writing proofs.
My attempt:
Assume to the contrary that solution of this equation is greater then or equal to 2.
Let $x=\frac 2p$.
Then we have
$ (\frac 2p)^5-2(\frac 2p)^3-3=0 $
$ \frac {2^5}{p^5} - \frac {2^4}{p^3} - 3 = 0$
We now consider two case: when $p=1$ and $p<1$.
when $p=1$:
$ 2^5 - 2^4 - 3 = 32 - 16 - 3 = 13$.
Since $x = 2$ is not a solution this is a contradiction.
When $p<1$:
If $p<1$, then we know $1/p>1.$ This implies $\frac {1}{p^{n+1}} > \frac {1}{p^n}.$
Since $\frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$, it is clear that the inequality
$ \frac 1{p^5}(2^5)- \frac 1{p^3} 2^4-3 > \frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$ holds
Since for all $x > 2$ is not a solution this is a contradiction.
Thus, solution $x$ is less then 2.

Let $x$ be a root and $x>2$.
Thus, $$x^5-2x^3-3=x^5-2x^4+2x^4-4x^3+2x^3-3>0,$$ which is a contradiction.
