Find all possible values of $c^2$ in a system of equations.

Question

Numbers $x,y,z,c\in \Bbb R$ satisfy the following system of equations: $$x(y+z)=20$$ $$y(z+x)=13$$ $$z(x+y)=c^2$$ Find all possible values of $c^2$.

To try to solve this, I expanded the equations: $$xy+xz=20$$ $$yz+xy=13$$ $$xz+yz=c^2$$ Then I subtracted the first equation from the second one to get: $$xz-yz=7$$ I added and subtracted this equation with the 3rd and got the following equations: $$2yz=7-c^2$$ $$2xz=7+c^2$$ I then added these equations, factored out $2z$ and divided by $2$ to get: $$z(x+y)=7$$ So, I found one possible value of $c^2 = 7$. How do I find the other values if they exist or how do I prove that there are no other values if they don't? Thanks!

Answer 1

On simplifying you will get

$yz=\frac{c^{2}-7}{2}$

$xz=\frac{c^{2}+7}{2}$

$xy=\frac{33-c^{2}}{2}$

Multiply above equations to get:

$x^{2}y^{2}z^{2}=\frac{c^{2}-7}{2}\frac{c^{2}+7}{2}\frac{33-c^{2}}{2}$

Since $x^{2}y^{2}z^{2} \geq 0$, therefore $7\leq c^2 \leq 33$. However, when you substitute it back to get $x,\ y,\ \& \ z$, you will find that $c^{2} \neq 7,\ 33$. Thus, possible value of $c^2$ are

$7 < c^2 < 33$

Answer 2

You just proved that it is necessary that $c^2 = 7$ if $z \neq 0$ (because you divided by $2z$). Now, you have to find a solution to the system (where $z\neq 0$) with $c^2 = 7$. If there is no such solution, there is no $c$ in this case. Also, consider the system with $z = 0$.

EDIT: I relied on what you wrote. When you subtract you do not have $2yz = 7-c^2$, but $2yz = c^2-7$. Sorry for the confusion. And yes, $c^2 = 7$ is not possible.

Answer 3

Let $c^2 = s$. Eliminating $x$ and $y$, you get an equation in $s$ and $z$: $$ s^2 + 2 z^2 s - 66 z^2 - 49 $$ Thus $$ z^2 = \dfrac{s^2 - 49}{66 - 2 s}$$ Since $z^2 \ge 0$, we need either $s \le -7$ or $7 \le s < 33$. This corresponds to $\sqrt{7} \le c < \sqrt{33}$. We then have $$ \eqalign{y &= \dfrac{z (33 - c^2)}{c^2 + 7}\cr x &= \dfrac{z(33 - c^2)}{c^2-7} \cr}$$ In particular, $c = \sqrt{7}$ is not possible. So the result is that $\sqrt{7} < c < \sqrt{33}$.