Find $x,y \in \mathbb C $ such that $x^5+y^5=275, x+y=5$.
By Matthew Cannon
My attempt:
Let $x, y$ be roots of$$t^2-5t+p=0$$I got$$p=\frac{5 \pm \sqrt{-431}}{2}$$using Vieta's relations. Now I just calculated x and y using the quadratic formula in $t^2-5t+p=0$. But I got two very hideous complex values of x and y. So I think my answer may be wrong.
The values I got are$$x= \frac{5+\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}$$ or$$x= \frac{5+\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}$$
Please post your own solutions too if you have any different solutions.
$\endgroup$ 33 Answers
$\begingroup$Observing that the pairs $(2,3)$ and $(3,2)$ are solutions and substituting $y=5-x$ from the second equation into the first one one obtains:$$ (x^2 - 5x + 19)(x - 2)(x - 3)=0. $$
Can you take it from here?
$\endgroup$ $\begingroup$Hint: Substituting $y=5-x$ the equation $x^5+y^5=275$ is equivalent to$$ (x^2 - 5x + 19)(x - 2)(x - 3)=0 $$
$\endgroup$ 2 $\begingroup$$$x^3+y^3=(x+y)^3-3xy(x+y)=5^3-3xy(5)$$
$$x^2+y^2=(x+y)^2-2xy=5^2-2xy$$
$$(x^3+y^3)(x^2+y^2)=x^5+y^5+x^2y^2(x+y)$$
$$\iff(125-15xy)(25-2xy)=275+5x^2y^2$$
$$\iff(xy)^2-25xy+114=0$$
For the rest, follow How to solve for $x$ in $\sqrt[4]{x+27}+\sqrt[4]{55-x}=4$?
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