MHB Find Solutions to a+b+c+d=4, a^2+b^2+c^2+d^2=6, a^3+b^3+c^3+d^3=94/9 in [0,2]

[anemone]

Find all the solutions to the system

$a+b+c+d=4\\a^2+b^2+c^2+d^2=6\\a^3+b^3+c^3+d^3=\dfrac{94}{9}$
in $[0, 2]$.

 

[Opalg]

Find all the solutions to the system $p_1 = a+b+c+d=4\\p_2 = a^2+b^2+c^2+d^2=6\\ p_3 = a^3+b^3+c^3+d^3=\dfrac{94}{9}$ in $[0, 2]$.

Spoiler

Let $x^4 - e_1x^3 + e_2x^2 - e_3x + e_4 = 0$ be the equation with roots $a,b,c,d$. By Newton's identities, $$\textstyle e_1 = p_1 = 4,\qquad e_2 = \frac12(p_1^2 - p_2) = \frac12(16 - 6) = 5,\qquad e_3 = \frac16(p_1^3 - 3p_1p_2 + 2p_3) = \frac16(64 - 72 + \frac{188}9) = \frac{58}{27}.$$ So the equation is $x^4 - 4x^3 + 5x^2 - \frac{58}{27}x + e_4 = 0$. Since $58$ is close to twice $27$, write the equation as $$x^4 - 4x^3 + 5x^2 - 2x = \tfrac4{27}x - e_4, \\ x(x-2)(x^2 - 2x + 1) = \tfrac4{27}(x-s),$$ where $s$ is a constant. Now look at the graph:

[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-0.47946479454949004,"ymin":-1.7930541038513184,"xmax":2.51919243201301,"ymax":1.9552674293518066}},"randomSeed":"5a5979bd9ad99a52fd1cfa25e8de4160","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"x^{4}-4x^{3}+5x^{2}-2x"},{"type":"expression","id":"2","color":"#2d70b3","latex":"\\frac{4}{27}\\left(x-s\\right)"},{"type":"expression","id":"4","color":"#6042a6","latex":"s=2","hidden":true,"slider":{"hardMin":true,"hardMax":true,"min":"1","max":"3","step":"0.1"}},{"type":"expression","id":"5","color":"#000000"}]}}[/DESMOS]
The roots of the equation are the points where the blue line meets the red curve. By using the slider, you can see that if $s<2$ then the largest root is greater than $2$. But if $s>2$ then the blue line goes lower, and only meets the red curve in two points, which means that two of the roots of the quartic equation are complex. So for the equation to have four real roots in the interval $[0,2]$, $s$ must be equal to $2$. After multiplying by $27$ the equation then becomes $(x-2)(27x^3 - 54x^2 + 27x - 4) = 0$, which factorises as $(3x-1)^2(3x-4)(x-2) = 0$. Therefore the solutions to the system are $\{a,b,c,d\} = \{\frac13,\frac13,\frac43,2\}$ (in any order).

 

[anemone]

Awesome, Opalg!(Cool) And thanks for participating!

Spoiler: Solution of other

I will start from the quartic equation $p(x)=x^4-4x^3+5x^2-\dfrac{58}{27}x+k$ where $p(x)$ has roots $a, b, c, d$.

$p'(x)=4x^3-12x^2+10x-\dfrac{58}{27}=\dfrac{2}{27}(3x-1)(18x^2-48x+29)$

Solving $p/(x)=0$ gives $x=\dfrac{1}{3},\,\dfrac{4}{3}\pm\dfrac{\sqrt{6}}{2}$.

Since $p(x)$ is a 4th degree polynomial with positive leading coefficient and $p'(x)$ has 3 distinct real roots in $(0, 2)$, it follows that in order for $a, b, c, d$ to be solutions of the given equations where $0\le a, b, c, d\le 2$, we must have

$p(0)\ge 0,\,p\left(\dfrac{1}{3}\right)\le0,\,p\left(\dfrac{4}{3}-\dfrac{\sqrt{6}}{2}\right)\ge0,\, p\left(\dfrac{4}{3}+\dfrac{\sqrt{6}}{2}\right)\le0,\,p(2)\ge 0$

Evaluating, we find $p\left(\dfrac{1}{3}\right)=p(2)=k-\dfrac{8}{27}$. Hence, $k=\dfrac{8}{27}$, from which we obtain

$\begin{align*}p(x)&=x^4-4x^3+5x^2-\dfrac{58}{27}x+\dfrac{8}{27}\\&=\dfrac{1}{27}(27x^4-108x^3+135x^2-58x+8)\\&=\dfrac{1}{27}(3x-1)^2(3x-4)(x-2)\end{align*}$

Therefore, the solutions in $[0, 2]$ are the 12 permutations of $\left(\dfrac{1}{3},\, \dfrac{1}{3},\, \dfrac{4}{3},\,2 \right)$.