While I was solving the question given by Harsh Kumar I could not find the real roots just because I changed the sign of $5y^2$. I also proved that there don't exist any real solution of that equation.

The equation I made by mistake was: $$5x^2+5y^2-8xy-2x-4y+5=0.$$

Please help me to find the solutions if exist.

  • $\begingroup$ Perhaps a more elegant proof would be by rewriting this as a quadratic form. That way, simply showing this to be an ellipsoid with negative square semiaxies is enough. $\endgroup$ – orion Dec 19 '16 at 13:03

It is clear that there exist no real solution for the equation $$5x^2+5y^2-8xy-2x-4y+5=0.$$

Rewriting the equation as a quadratic in $y$ $$5y^2-y(8x+4)+5x^2-2x+5=0$$ on using quadratic formula $$y=\frac{(8x+4)\pm2\sqrt{-9x^2+26x-21}}{10}\tag{1}$$

Now the expression under the radical sign in $(1)$ can never be positive and therefore $y$ cannot take a real value for any real value of $x$. In fact $$-9x^2+26x-21=-9\times\left[\left(x-\frac{13}{9}\right)^{2}+\frac{20}{81}\right]$$

which is negative, whatever the value of $x\in\Bbb R$ may have.

Hence, the equation $5x^2+5y^2-8xy-2x-4y+5=0$ can not be satisfied by any pair of real numbers $x$ and $y$.


We may rewrite this as $$ 4x^2 - 8xy + 4y^2 + x^2 -2x + 1 + y^2 - 4y + 4 = 0\\ (2x - 2y)^2 + (x-1)^2 + (y-2)^2 = 0 $$ and we see that this is a sum of three (real) squares that equals $0$. If the sum of three squares equals $0$, then that means that each of the squares are zero. But $(x-1)^2 = 0$ means $x = 1$ and $(y-2)^2 = 0$ means $y = 2$, and then $(2x-2y)^2 \neq 0$. Therefore there can be no real solutions to this polynomial equation.

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    $\begingroup$ It should be $(2x-2y)^2$ not $(2x+2y)^2$ $\endgroup$ – i9Fn Dec 28 '16 at 13:54
  • $\begingroup$ @i9Fn You're right. Fixed it. $\endgroup$ – Arthur Dec 28 '16 at 18:07

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