Solutions and permutations

Algebra Level 5

$\left\{ \begin{array}{lcc} x^2+y^2+z^2 & = & 50 \\ (x+y)(y+z)(z+x) & = & 14 \\ x^3+y^3+z^3-3xyz& = & 146 \end{array} \right.$

How many rational solutions does the system of equations above have?

The answer is 6.

1 solution

Paola Ramírez
Apr 23, 2016

Let be $a=x+y+z, b=xy+xz+yz, c=xyz$ . By factoring identities, the system is written a

$\left\{ \begin{array}{lcc} a^2-2b & = & 50 \\ ab-c & = & 14 \\ a^3-3ab& = & 146 \end{array} \right.$

Then, we get that $a^3-150a+292=(a-2)(a^2+2 a-146)=0$ . As we are looking for real solutions, $p=2$ so $q=-23$ and $r=-60$ .

Finally, by Vieta's Formulas, $x,y,z$ are roots of the equation $d^3-2d^2-23d+60 \therefore$ the number of solutions is any permutation of $3,4,-5$ , in other words $\boxed{6\text{ solutions}}$

Buen uso de las identidades, incluso usaste las mismas letras que yo. (+1)

Alan Enrique Ontiveros Salazar - 5 years, 1 month ago

I don't know Spanish (If that's what it is): But lemme give a shot at translating it:

Good use of Identities, used the same letters (+1) (maybe)

Can you please verify? :P

Mehul Arora - 5 years ago

Yes, it's basically the same, I would translate it as "Good use of the identities, even you used the same letters than me".

Alan Enrique Ontiveros Salazar - 5 years ago

@Alan Enrique Ontiveros Salazar – Oh yay! :P

Mehul Arora - 5 years ago

Solutions and permutations

The answer is 6.

1 solution

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