Condition for $x^4-18x^2+4dx+9=0$ has four real roots Prove that if $x^4-18x^2+4dx+9=0$ has four real roots, Then $d^4 \le 1728$
My try:
obviously the equation should have two unequal negative roots and two repeated positive roots OR vice versa.
Choosing first case here is the rough sketch:

Let the two negative roots are $-p_1, -p_2$ and repeated positive root be $p_3$
where $p_1,p_2,p_3 \gt 0$
Thus we have the polynomial identity as:
$$(x+p_1)(x+p_2)(x-p_3)^2=x^4-18x^2+4dx+9$$
Comparing coefficients we get:
$$p_1+p_2=2p_3 \tag{1}$$
$$p_1p_2=3p_3^2-18\tag{2}$$
$$p_1p_2p_3^2=9\tag{3}$$
$$p_3^2(p_1+p_2)-2p_1p_2p_3=4d\tag{4}$$
From $(2)$ and$ (3)$ we get:
$$\frac{9}{p_3^2}=3p_3^2-18$$
we get:
$$p_3^4-6p_3^2-3=0$$
Since $p_3 \gt 0$ we get
$$p_3=\sqrt{3+2\sqrt{3}}$$
Using $(4)$ we get:
$$2p_3^3-\frac{18}{p_3}=4d$$
From this we get a unique value of $d$
But how proof is to be done, what went wrong in my analysis? 

Here is an alternative approach to the original question. As you have reasoned, it is helpful to solve for the $d$ that make $f(x)=x^4-18x^2+4dx+9$ have a repeated root. If $f$ has a repeated root, then the gcd of $f$ and $f'$ also shares that root. 
We have $f'(x)=4x^3-36x+4d=4\left(x^3-9x+d\right)$.
We look for a linear combination of $f$ and $f'$ that is just a linear polynomial. Starting with $$\left(-x^2+Ax+B\right)\frac{f'(x)}{4}+\left(x+C\right)f(x)=mx+b$$ 
we slowly deduce values for $A,B,C$ and find
$$\left(-x^2-\frac{d}{3}x+9\right)\left(x^3-9x+d\right)+\left(x+\frac{d}{3}\right)\left(x^4-18x^2+4dx+9\right)=\left(d^2-72\right)x+12d$$
Since we are assuming there is a repeated root, it must be the root of this linear polynomial, $\frac{12d}{72-d^2}$.
So $\frac{1}{4}f'\mathopen{}\left(\frac{12d}{72-d^2}\right)\mathclose=0$.
$$\begin{align}
\left(\frac{12d}{72-d^2}\right)^3-9\left(\frac{12d}{72-d^2}\right)+d&=0\\
(12d)^3-108d\left(72-d^2\right)^2+d\left(72-d^2\right)^3&=0\\
-d\left(d^6-108d^4-1728d^2+186624\right)&=0\\
-d\left(d^2-108\right)\left(d^4-1728\right)&=0
\end{align}$$
So (assuming we are searching for the positive $d$) either $d=\sqrt{108}$ or $d=\sqrt[4]{1728}$. The first would mean $-\sqrt{12}$ is the common root of $f$ and $f'$. But try this for $x$ in $f(x)$ (with $d=\sqrt{108}$) and it does not give $0$.
Therefore $d=\sqrt[4]{1728}$. From here you already have the reasoning for why $\lvert d\rvert\leq\sqrt[4]{1728}$ is the general condition for four real roots.
