<s>[INST] Find $\mathbf a$ for which the following equation has triple roots Find $\mathbf a \in\mathbb{R}$ for which the following equation has triple roots:  
$$ x^4-5x^3+\mathbf ax^2-7x+2=0$$
 [/INST] Suppose $b$ is the triple root. Then the polynomial can be factored as $(x-b)^3(x-c)$ and we get
$$
x^4 - (3b+c)x^3 + (3b^2 + 3bc)x^2 - (b^3+3b^2c)x + b^3c
$$
so we need
\begin{cases}
3b+c=5 \\
3b^2+3bc=a\\
b^3+3b^2c=7\\
b^3c=2
\end{cases}
Multiplying the third equation by $b$ and taking into account the fourth equation we get
$$
b^4-7b+6=0
$$
The fourth equation, together with the first, becomes $5b^3-3b^4=2$
$$
3b^4-5b^3+2=0
$$
Obviously $b=1$ is a solution for both. If $b\ne1$ we get
\begin{cases}
b^3+b^2+b-6=0 \\
3b^3-2b^2-2b-2=0
\end{cases}
and, eliminating $b^3$,
$$
5b^2+5b-16=0
$$
Multiplying the top equation by $2$ and summing up,
$$
5b^3-14=0
$$
Since $5b^3+5b^2-16b=0$, we get $5b^2-16b+14=0$, that finally yields $5b-16=-16b+14$, or $b=10/7$, that doesn't satisfy $5b^3=14$.
Therefore $b=1$ and $c=2$, so $a=9$.
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