Given $x^2 - y^2 = 9$ and $xy = 3$. Find $x + y$
Matthew Harrington 
Hi can someone help me with this?
Given $x^2 - y^2 = 9$ and $xy = 3$. Find $x + y$
My solution:\begin{align*} x^2 - y^2 = 9 & \Longleftrightarrow (x^2 - y^2)^2 = 81\\\\ & \Longleftrightarrow x^4 - 2x^2y^2 + y^4 = 81\\\\ & \Longleftrightarrow x^4 + y^4 = 99\\\\ & \Longleftrightarrow (x^2 + y^2)(x^2 - y^2) = 99 \end{align*}
Since $x^2 - y^2 = 9$, it results that $9(x^2 + y^2) = 99$ and $x^2 + y^2 = 11$.
Gathering all previous results, one has that\begin{align*} 2x^2 = 20 & \Longleftrightarrow (x = \sqrt{10})\wedge(y = 1) \end{align*}
The answer on this site is different from mine :
Could someone pls explain this? Thanks
$\endgroup$ 33 Answers
$\begingroup$We have the polynomial identity $$(x+y)^4=(x^2-y^2)^2+4 x y (x+y)^2$$
Call $z = (x+y)^2$ and use $x^2-y^2 =9$, $x y=3$ to get$$z^2=9^2+4\cdot 3\,z$$
Solve for $z$ to get $$z = \frac{12\pm\sqrt{12^2+4\cdot 81}}{2} = 6 \pm 3\sqrt{13}$$
Discard the negative solution since $z = (x+y)^2 \ge 0$, so $z = 6+3\sqrt{13}$.
Then $x+y = \pm \sqrt{z} = \pm\sqrt{6+3\sqrt{13}}$
$\endgroup$ 1 $\begingroup$HINT
Here is another way to approach it for the sake of curiosity.
Since $xy = 3$, we conclude that $x\neq 0$ and $y\neq 0$. Based on such relations, we can divide both sides of $x^{2} - y^{2} = 9$ by $xy$ and make the change of variables $t = \dfrac{x}{y}$:\begin{align*} \frac{x^{2} - y^{2}}{xy} = \frac{9}{3} & \Longleftrightarrow \frac{x}{y} - \frac{y}{x} = 3\\\\ & \Longleftrightarrow t - \frac{1}{t} = 3\\\\ & \Longleftrightarrow t^{2} - 3t - 1 = 0\\\\ & \Longleftrightarrow t = \frac{3 \pm \sqrt{13}}{2} \end{align*}
Consequently, one has the following collection of systems of equations to solve:\begin{align*} \begin{cases} xy = 3\\\\ \dfrac{x}{y} = \dfrac{3 + \sqrt{13}}{2} \end{cases}\quad \begin{cases} xy = 3\\\\ \dfrac{x}{y} = \dfrac{3 - \sqrt{13}}{2} \end{cases} \end{align*}
Can you take it from here?
EDIT
Observe the second system of equations has no solution, because $xy = 3 > 0$, and $\dfrac{x}{y} < 0$.
Hence it suffices to consider the first system of equations.
If we multiply both of them together, we get\begin{align*} x^{2} = \frac{9 + 3\sqrt{13}}{2} & \Longleftrightarrow x = \pm\sqrt{\frac{9 + 3\sqrt{13}}{2}}\\\\ & \Longleftrightarrow y = \pm\sqrt{\frac{-9 +3\sqrt{13}}{2}} \end{align*}
$\endgroup$ 5 $\begingroup$Here's what I think is a more straightforward approach than the video, but it gives the solution in a different form. Substitute $y=3/x$ to obtain $x^2-(3/x)^2=9$. Multiply both sides by $x^2$, yielding $x^4-9=9x^2$. Rewrite as $(x^2)^2-9x^2-9=0$, and apply the quadratic formula to obtain$$x^2 = \frac{9\pm3\sqrt{13}}{2}.$$Because $x^2 \ge 0$, you can discard the extraneous solution, leaving$$x^2 = \frac{9+3\sqrt{13}}{2},$$which implies that$$x = \pm \sqrt{\frac{9+3\sqrt{13}}{2}}.$$Now $$y = \frac{3}{x} = \pm \frac{3}{\sqrt{\frac{9+3\sqrt{13}}{2}}} = \pm \sqrt{\frac{6}{3+\sqrt{13}}} \cdot \sqrt{\frac{3-\sqrt{13}}{3-\sqrt{13}}} = \pm \sqrt{\frac{6(3-\sqrt{13})}{9-13}} = \pm \sqrt{\frac{-9+3\sqrt{13}}{2}} ,$$and so$$x+y=\pm \left(\sqrt{\frac{9+3\sqrt{13}}{2}}+\sqrt{\frac{-9+3\sqrt{13}}{2}}\right).$$
To see that this solution matches the video, you can square both sides to obtain\begin{align} (x+y)^2 &= \left(\sqrt{\frac{9+3\sqrt{13}}{2}}+\sqrt{\frac{-9+3\sqrt{13}}{2}}\right)^2 \\ &= \frac{9+3\sqrt{13}}{2} + \frac{-9+3\sqrt{13}}{2} + 2\sqrt{\frac{9+3\sqrt{13}}{2}}\sqrt{\frac{-9+3\sqrt{13}}{2}} \\ &= 3\sqrt{13} + \sqrt{(9+3\sqrt{13})(-9+3\sqrt{13})} \\ &= 3\sqrt{13} + \sqrt{-81+117} \\ &= 3\sqrt{13} + 6, \end{align}so$$x+y = \pm \sqrt{6+3\sqrt{13}}.$$
$\endgroup$ 1
