Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$ in real numbers

Question

Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$

The problem does not say it but I think solutions should be from $\mathbb{R}$. I tried to express the left sum as a sum of squares but that does not work out. Any suggestions?

Answer 1

$$2x^2 + 5y^2 + 6xy -2x -4y+1=0$$

$$(1+1)x^2 + (4+1)y^2 + (4+2)xy - 2x -4y + 1=0$$

$$(x^2 +4y^2 +4xy -2x -4y + 1) + (x^2 +2xy + y^2)=0$$

$$(x+2y-1)^2 + (x+y)^2=0$$

Your idea to write the expression as a sum of squares is good intuition. Now, when can a sum of squares be zero? Exactly when both of the squares are zero.

Thus

$$x+2y-1=0$$

$$x+y=0$$

You should be able to find the solution from here.

Answer 2

You can solve for $x$:

$(2)x^2+(6y-2)x+(5y^2-4y+1)=0\implies$

$x_{1,2}=\frac{-(6y-2)\pm\sqrt{(6y-2)^2-4\cdot2\cdot(5y^2-4y+1)}}{2\cdot2}=\frac{-6y+2\pm\sqrt{-4y^2+8y-4}}{4}=\frac{-6y+2\pm\sqrt{-4(y-1)^2}}{4}=\frac{-6y+2\pm2i(y-1)}{4}$

Then the only real solution is with $y=1$, hence $x=-1$.

Answer 3

You can consider that the equation is quadratic in $y$ and then the solution is given by $$y=\frac{1}{5} \left(2-3 x \pm\sqrt{-(x+1)^2}\right)$$ In the real domain, $x=-1$ (because of the radical) and then $y=1$.