Proving that a transformation of a function gives a positive result

Question

If $x$ is real and: $$p = \frac{3(x^2+1)}{2x-1}$$

Prove that: $$ p^2-3(p+3)\geq 0$$

I think this has something to do with equating the discriminant to $0$, but I'm not entirely sure

I'd really appreciate any help at all on this. Thanks!

Answer 1

The first equation can be rewritten as$$3x^2-2px+p+3=0.$$ Now, we have a quadratic equation $ax^2+bx+c=0$ with $a=3$, $b=-2p$, and $c=p+3$. The solution(s) of the equation are given by the formula $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2}.$$ The solution(s) are real iff the expression under the square root sign is positive (or zero). So the condition for the discriminant is: $$b^2-4ac=4p^2-4\times3\times(p+3)\ge0.$$ Dividing by $4$ we get $$p^2-3(p+3)\ge 0$$ then, and only then, the solution(s) of the first equation will be real.