Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1)  \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I can see that $p^2 > -3(p+3)$ for  almost all values of $p \ $, therefore
$p^2-3(p+3) > 0 $. However, the question asks to show that $\ p^2-3(p+3) \geq 0$  
If I make $p^2 = 3(p+3)$ I can find roots and so $\ p^2-3(p+3) = 0$, when $\displaystyle{p = \frac{3 \pm \sqrt{45}}{2}}$. Therefore  $\ p^2-3(p+3) \geq 0$  
Having done this, how can I mathematically show that $p^2$ is never $<$ than $3(p+3)$? Because I am not satisfied with just saying that by inspection $p^2$ is greater than $3(p+3)$.
Thank you

You have $$3x^2 - 2px + (p+3) = 0.$$  Given that $x$ is real the quadratic needs to have a discriminant $\Delta \ge 0$. So $$\Delta = 4p^2 - 12(p+3) \geq 0.$$
Dividing by $4$ yields $$\bbox[10px, border: blue 1px solid]{p^2 - 3(p+3) \ge 0.}$$ as required.
