Determine $x,y,z ∈ R$ such that $2x^2+y^2+2z^2-8x+2y-2xy+2xz-16z+35=0$

Question

I have got a question which is as follow:

Determine $x,y,z ∈ R$ such that $2x^2+y^2+2z^2-8x+2y-2xy+2xz-16z+35=0$.

This is a quite filthy looking equation and I know that solving it must involve some factorisation etc, but the problem is my thoughts are not seeming to work on paper.

I need your help. Thanks in advance.

Answer 1

It is just the matter of rearrangement. You have got it right, it will require some factorisation (maybe not some). But how?? Let's see:

$$2x^2+y^2+2z^2-8x+2y-2xy+2xz-16z+35=0$$

$$\implies (x^2+y^2-2xy) + (x^2+z^2+2xz) + z^2-16z-8x+2y+35=0$$

$$\implies (x-y)^2 +(x+z)^2 + (z^2-10z+25)-6z-8x+2y+10=0$$

$$\implies ((x-y)^2-2(x-y)+1) + ((x+z)^2-6(x+z)+9)+(z^2-10z+25)=0$$

$$\implies (x-y-1)^2 + (x+z-3)^2+(z-5)^2=0$$

The only way in which all these perfect square can give a sum of zero is that when all of these are zero (because $x,y,z$ are real not complex). So, $$x-y-1=0, x+z=3 , z=5$$

On solving these three you will get that $(x,y,z)=\;(-2,-3,5)\;$. We are done I think :) :)

Answer 2

Following lab bhattacharjee's hint ...

Let $f(x,y,z) = 2x^2 + y^2 + 2z^2 - 8x + 2y - 2xy + 2xz - 16z + 35$.

Expressing $f(x,y,z)$ in the form $$2z^2 + (2x - 16)z + (2x^2-2xy + y^2-8x + 2y + 35) = 0$$ as a quadratic in $z$, the discriminant is $$4(-3x^2 + 4xy - 2y^2 -4y - 6)$$ hence, for fixed real numbers $x,y$, there is a real number $z$ such that $f(x,y,z) = 0$ if and only if $g(x,y) \ge 0$, where $$g(x,y) = -3x^2 + 4xy - 2y^2 -4y - 6$$ Expressing $g(x,y)$ in the form $$-2y^2 + (4x-4)y - (3x^2 + 6)$$ as a quadratic in $y$, the discriminant is $$-8(x+2)^2$$ hence, $g(x,y) \ge 0$ is solvable for real $x,y$ if and only if $x=-2$.

Then $g(-2,y) = -2(y+3)^2$, hence $g(x,y) \ge 0\;$has the unique solution $(x,y) = (-2,-3)$.

Finally, for $x = -2,y = -3$, $$f(-2,-3,z) = 0 \iff 2(z - 5)^2 = 0 \iff z = 5$$ Thus, the the equation $f(x,y,z) = 0$ is satisfied by exactly one real triple, namely $(x,y,z) = (-2,-3,5)$.