Find $\mathbf a$ for which the following equation has triple roots

Question

Find $\mathbf a \in\mathbb{R}$ for which the following equation has triple roots:

$$ x^4-5x^3+\mathbf ax^2-7x+2=0$$

Answer 1

Let $r_1$ be the triple root. Then there must be an additional single root $r_2$ to match the fourth degree equation.

Then from the Vieta formulas:

$3r_1+r_2=5$

$r_1^3r_2=2$

Eliminating $r_2$ from this pair:

$r_1^3(5-3r_1)=2$

$3r_1^4-5r_1^3+2=0$

This has multiple roots because there are different choices for the triple root. Here we shall take the obvious rational root $r_1=1$, whereupon $r_2=2$.

Then the polynomial must factor as $(x-1)^3(x-2)$ and $a$ is determined accordingly.

Answer 2

A triple root of $f$ is also a root of $f''$, and so is a root of their resultant, which is $$400 a^4 - 2400 a^3 - 82080 a^2 + 892944 a - 2262816$$

It so happens that $a=9$ is a root (and the only nice one), but let's avoid guessing.

A triple root of $f$ is also a root of $f'$, and so is a root of their resultant, which is $$32 a^4 - 396 a^3 - 4887 a^2 + 90882 a - 343359 $$ Now, the gcd of these two polynomials in $a$ is $a-9$ and so only $a=9$ works.

Answer 3

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 $a=9$.