Nature of the roots of quadratic equation

Question

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

Answer 1

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.

Answer 2

It must be $b^2-4ac\geq 0$ because it is given $x$ is real so equation must have real roots which may be either equal or unequal

Answer 3

$$3x^2-2px+(p+3) = 0 \\ 9x^2-6px+3(p+3) = 0 \\ (3x-p)^2+(3(p+3)-p^2)=0 \\ p^2-3(p+3) = (3x-p)^2 \ge 0$$

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Furthermore you can see that $p^2-3(p+3) = 0$ when $p=3x$ and $x^2 = x+1$, i.e. $x = \frac{1 \pm \sqrt 5}2$

Answer 4

I think you are overthinking this.

To show that $x$ satisfies the quadratic equation $3x^2−2px+(p+3)=0$, you have two cases :

(1) $b^2−4ac > 0$ :

Therefore $4(p^2−3(p+3)) > 0$ i.e. $p^2−3(p+3) > 0$

(2) $b^2−4ac = 0$ :

Then $4(p^2−3(p+3)) = 0$ i.e. $p^2−3(p+3) = 0$

The third case $b^2−4ac < 0$ here is irrevelant since $x$ is supposed to be a solution.

No need to compute anything here.