# System of quadratic equations with three variables

The problem is as follows:

For $x,y,z \in R$, $$\left\{ \begin{array}{l} x^{2} -yz-8x+7=0 \\ y^{2}+z^{2}+yz-6x+6=0 \end{array} \right.$$ What is the domain of $x$?

One way to solve this is to use another variable. This is shown in this answer. What other ways are there to solve this?

Addition: The answer to this question is $1 \leq x \leq 9$.

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Would you please show the change of variable that can be used to solve this problem -- thanks. – user84413 Jul 17 '14 at 14:52
-1 because you don't show us your solution – miracle173 Jul 19 '14 at 19:02
@miracle173, you will find the solution if you scroll down a bit. I didn't want to include it in the question because someone might want to solve this before seeing the solution. – mentoR Jul 19 '14 at 19:06
+1 sorry, I did not see this. I had to change your post otherwise I could not change the downvote to an upvote. But I hope that my edit makes sense anyway – miracle173 Jul 19 '14 at 19:17
Why didn't I think of that? @miracle173 thanks for the edit. – mentoR Jul 19 '14 at 19:20

We have that $yz=x^2-8x+7$ and, substituting into the 2nd equation, $y^2+z^2=6x-6-yz=6x-6-(x^2-8x+7)=-x^2+14x-13$.

Since the system of equations $y^2+z^2=a$ and $yz=b$ has a solution iff $a\ge2b$ and $a\ge-2b$,

as shown below,

the given system has a solution iff

1) $-x^2+14x-13\ge2(x^2-8x+7)$ and 2) $-x^2+14x-13\ge -2(x^2-8x+7)$.

Since 1) $-x^2+14x-13\ge2(x^2-8x+7) \iff -3x^2+30x-27\ge0 \iff x^2-10x+9\le0 \iff (x-1)(x-9)\le0 \iff 1\le x\le9$

and 2) $-x^2+14x-13\ge -2(x^2-8x+7) \iff x^2-2x+1\ge0 \iff (x-1)^2\ge0$, which is true $\;\;\;\;\;\;\;\;\;$for all $x\in\mathbb{R}$,

the values of $x$ for which this system has a solution are the values of $x$ in $[1,9]$.

$----------------------------------------$

$\Longrightarrow$ If $y^2+z^2=a$ and $yz=b$ has a solution, then $y^2-2yz+z^2=(y-z)^2\ge0\implies a\ge2b$, and $y^2+2yz+z^2=(y+z)^2\ge0\implies a\ge-2b$.

$\Longleftarrow$ If $a\ge2b$ and $a\ge-2b$, then $a\ge2|b|\ge0$ and $a^2\ge4b^2$.

If $a=0$, then $b=0$ and the system has the solution $y=0, z=0$.

If $a>0$, then $y=\big(\frac{a+\sqrt{a^2-4b^2}}{2}\big)^{1/2}$ satisfies $y^4-ay^2+b^2=0$, so $y^2-a+\frac{b^2}{y^2}=0$ and therefore letting $z=\frac{b}{y}$ gives a solution of $y^2+z^2=a$ and $yz=b$.

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A partial solution of the problem :

The sum of the two equations (which must hold both) is :

$$x^2+y^2+z^2-14x+13=0$$

Completing the square yields

$$x^2-14x+49+y^2+z^2=-13+49=36$$

So

$$(x-7)^2+y^2+z^2=36$$

So, all the points of the relation must be on the sphere with center (7,0,0) and radius 6.

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Thank you for the quick response. But it's not correct. I'll edit my question a bit. – mentoR Jul 7 '14 at 2:05
On a second thought, did you mean a necessary condition by a 'partial' solution? Then, you're right. But my question wasn't about the answer to the problem but rather a different way of solving the problem. – mentoR Jul 7 '14 at 2:16

I'm adding a solution to the question in response to the request of user84413. And I'm hoping for someone to answer the question from yet another approach. ;)

We can change the two equations as follows: $$\left\{ \begin{array}{l} yz=x^{2}-8x+7 =(x-1)(x-7) \\ (y+z)^{2}=yz+6x-6=(x-1)^{2} \longrightarrow y+z=\pm (x-1) \end{array} \right.$$

Now that we have the sum and the product of two numbers, we can write a quadratic equation with those two roots.

$$t^{2} \pm (x-1)t+(x-1)(x-7)=0$$

Since the two roots are real numbers, the discriminant should be non-negative.

$$D=(x-1)^{2}-4(x-1)(x-7) \geq 0$$

From this inequality, we get $1 \leq x \leq 9$.

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FYI, this solution is not my idea. – mentoR Jul 19 '14 at 19:08