<s>[INST] 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? 
 [/INST] 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.
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