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- #1

- Feb 14, 2012

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$a+b+c+d=4\\a^2+b^2+c^2+d^2=6\\a^3+b^3+c^3+d^3=\dfrac{94}{9}$

in $[0, 2]$.

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- Thread starter
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- #1

- Feb 14, 2012

- 3,522

$a+b+c+d=4\\a^2+b^2+c^2+d^2=6\\a^3+b^3+c^3+d^3=\dfrac{94}{9}$

in $[0, 2]$.

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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 $[0, 2]$.

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).

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- Feb 14, 2012

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

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 $\left(\dfrac{1}{3},\, \dfrac{1}{3},\, \dfrac{4}{3},\,2 \right)$.