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Im currently doing my Kumon (A math tutoring center I guess) homework, and Im having a bit of difficulty answering a simultaneous equation, involving $x$ and $y$ variables to the second power. School curriculum wise, we're not close to learning this, so I apologize if I don't understand something.

Here is the equation:

$\displaystyle xy = 12$

$\displaystyle x^2+y^2 = 25$

Assuming that $x+y = A$ and $xy = B$

If I'm forgetting something please let me know, as I'm kind of a noob at this. Any help is welcome!

Thanks!

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    $\begingroup$ The answers show how to do the solution in general. In this specific case, when you see $x^2+y^2=25=5^2$ you should immediately think $3,4,5$ triangle and note that $3 \cdot 4=12$. Done. School problems often are made to have easy solutions. $\endgroup$ – Ross Millikan Jun 27 '15 at 14:38
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Using $$x^2+y^2=x^2+y^2+2xy-2xy=(x+y)^2-2xy,$$ we have $$25=(x+y)^2-2\cdot 12\iff (x+y)^2=49\iff x+y=\pm 7.$$

So, we have $$(x+y,xy)=(7,12),(-7,12).$$

Can you take it from here?

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If we have $\displaystyle xy = 12 \implies y = \frac{12}{x}$ then we can substitute this into the second equation, yielding $$x^2 + \left(\frac{12}{x}\right)^2 = 25 \iff x^4 -25x^2 + 144 = 0$$

Which is a quadratic in $x^2$, $(x^2 - 9)(x^2 - 16) = 0$ with solutions $x = 3,4 -3, -4$.

The corresponding solutions for $y$ are $y = \frac{12}{3}, \frac{12}{4}, \frac{-12}{3}, \frac{-12}{4}$.

So our solutions are $(x,y)$: $(4,3)$, $(3,4)$, $(-3, -4)$, $(-4,-3)$

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Hint:

$$x^2(x^2+y^2)=x^4+x^2y^2=x^4+12^2=25x^2.$$

You should be able to solve the last equation, of the biquadratic type.


Advanced Hint:

Seeing a biquadratic equation tells you that there are up to four solutions, coming in two pairs of opposite signs. This was to be expected because swapping $x$ and $y$ doesn't change the equations, and changing the signs of $x$ and $y$ simultaneously doesn't change the equations.

Now school problems often have easy solution such as integer ones. So you can look for the ways to decompose $12$ in two factors (like $2\times6$ or $3\times4$), or decompose $25$ in two squares (like $16+9$).


Obviously, $(4,3)$ is a solution. The remaining three must be $(3,4), (-4,-3),(-3,-4)$.

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We have $x^2+2xy+y^2=(x+y)^2=49$.It follows that $x+y=\pm 7.$ In similar way $x^2-2xy+y^2=(x-y)^2=1$ or $x+y=\pm 1.$ So \begin{cases} x+y=\pm7,\\ x-y=\pm1. \end{cases} The solutions of the system are $$ x=\frac{1}{2}(\pm7\pm1), y=\frac{1}{2}(\mp1\pm7). $$ Taking the 4 different combination of signs we get the 4 solutions.

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