Finding exactly one real solution to the system

Question

I dont know how to go about finding the one real solution to the system where k is a real number

Thank you

Answer 1

You can write $n=\frac{6}{m}$. Substituting in the first equation, we get $m+\frac{6}{m}=k$, and then $m^2-km+6=0$.

This is a quadratic equation, which we can solve using the Quadratic Formula. We get $$m=\frac{k\pm\sqrt{k^2-24}}{2}.$$ There is a single solution if $k^2-24=0$. There is no real solution if $k^2-24\lt 0$.

Another way: Note that $(m-n)^2=(m+n)^2-4mn=k^2-24$. There is no real solution if $k^2-24\lt 0$, since the square of a real number cannot be negative.

If $k^2-24\gt 0$, we can take the square roots, and we get $m-n=\pm\sqrt{k^2-24}$. If $k^2-24=0$, we get $m=n$, and a single solution. If $k^2-24\gt 0$, there are two possible values of $m-n$, and therefore two possible values of $(m,n)$.

Remark: We can do the exactly one part with less computation. Note that the system of equations is symmetric in $m$ and $n$. Therefore if $(x,y)$ is a solution, so is $(y,x)$. That gives two solutions unless $x=y$. So our only chance at one solution is if $x=y$. That gives $x=y=\pm\sqrt{6}$, and therefore $k=x+y=\pm2\sqrt{6}$.