# Solving a system of quadratic equations.

Solve for real $(a, b, c)$ satisfying

$$ab + bc + ca = 1$$

$$a^2 − 2b^2 = 1$$

$$2b^2 − 3c^2 = 1$$

I try isolating $a$, but it leads to a very complicated expression in $a$.

-
I edited your question inserting LaTeX so that the mathematics is more visually appealing. You can click "edit" to look at the new code, and to see how this was done, or search how to use basic LaTeX. – Eric Naslund Dec 21 '12 at 0:03

If one can show that $c=0$ then $a=\sqrt{2},b=\sqrt{2}/2$ as noted in Tim's answer. I found it a bit involved to actually prove $c=0$ is a consequence of the equations. (I'd be interested in a simpler proof of $c=0$ than the following.) Label the equations:

[1] $ab+bc+ca=1,$

[2] $a^2-2b^2=1,$

[3] $2b^2-3c^2=1.$

From [2] and [3] we can see neither of $a,b$ are zero, since we seek real solutions. From [2] alone we can also see that neither of $a+b,a-b$ are zero, otherwise [2] says $-t^2=1$. We'll need these nonzero properties later.

Now since the right sides are all equal, we have from [1] and [2] that $$c(a+b)+ab=a^2-2b^2,$$ and moving the $ab$ to the right side and factoring gives $$c(a+b)=(a+b)(a-2b).$$ As noted we know $a+b$ is nonzero, so we now have $c$ in terms of $a,b$ as $$c=a-2b.$$ We next use that [2] and [3] have the same right sides, so that on replacing $c$ as above, $$a^2-2b^2=2b^2-3(a-2b)^2,$$ and the difference factors to obtain: $$4(a-2b)(a-b)=0.$$ Having noted that we know $a-b$ is not zero we arrive at $a-2b=0$, i.e. $c=0$ as desired.

EDIT: I removed a phrase "luckily the $b^2$ terms cancel" since they didn't. Also as others have noted, in case $c=0$ the signs on $a,b$ may be changed for another solution, so that the system has actually two triples $(a,b,c)$ for solutions.

-
There is one more solution when $c = 0$: the negative roots. That is, $a = - \sqrt{2}$ and $b = - \frac{\sqrt{2}}{2}$ – Paresh Dec 22 '12 at 9:45
Yes, that is also a solution. I was more concerned with showing $c=0$ followed from the equations, and so overlooked the sign change invariance. +1 on comment. – coffeemath Dec 23 '12 at 16:53

As one approach try putting the first equation $ab+bc+ca=1$ in terms of $c$ then set $c$ equal to $0$. It will look kind of complicated as you put it with variable terms under radical signs but you will see that the result equals $1$. From there the other two solutions for $a$ and $b$ will be apparent. Real solutions are $$a=\sqrt{2}$$ $$b=\frac{\sqrt{2}}{2}$$ $$c=0$$ to check your work.$$ab=\frac{\sqrt{18c^4 +18c^2+4}}{2}$$

$$bc=\frac{c\sqrt{2+6c^2}}{2}$$

$$ca=\frac{2c\sqrt{2+3c^2}}{2}$$

-
How do you justify setting $c$ equal to $0$? – Gerry Myerson Dec 22 '12 at 6:26
@GerryMyerson Because I knew that $ab+bc+ca=1$, a way to get $1$ was to set $c$ equal to $0$ to get $\frac{\sqrt{4}}{2}$ for $ab$ and since $c=0$ $bc$ and $ca$ also equal $0$ to get $1+0+0=1$. Now, $a=\pm\sqrt{2+3c^2}$ and $b=\pm\frac{\sqrt{2+6c^2}}{2}$. Having established that $c=0$ plugging $0$ for $c$ in the equations for $a$ and $b$ gives solutions for $a$ and $b$. That is how I got my solutions although I missed the $\pm$ for $a$ and $b$ in my original answer. – Tim Monahan Dec 22 '12 at 18:51
I don't follow. Yes, if you set $c=0$ you get some solutions. But maybe there are other values of $c$ that will lead to solutions. I don't see where you have "established that $c=0$". – Gerry Myerson Dec 23 '12 at 4:59
I suppose you are right that there could be other solutions. I did not try to find any other solutions. Do you know of other real solutions to this problem? I will have to think about this to see if there are other ways to prove that $c$ could only be $0$ in addition to the way coffeemath proved it. – Tim Monahan Dec 23 '12 at 7:26
If coffee's work is correct, then $c$ must be zero and there are no other real solutions (up to sign). – Gerry Myerson Dec 23 '12 at 16:33