Is there exist any real solution of the equation…

While solving a question given by my friend I found a problem in finding the real roots, I tried the question many times but I could not find any real solution of that equation.

The equation was:

$5x^2-5y^2-8xy-2x-4y+5=0$

• Two solutions are $x=0, y=-\frac25 \pm\frac15\sqrt{29}$. This is an equation in two variables, so solutions are going to be ordered pairs of numbers. I found this by setting $x$ equal to zero and solving the resulting quadratic in $y$. There are many more solutions, of course. – MPW Dec 15 '16 at 16:25
• Looks like the solutions form a hyperbola. Do you know how to diagonalize a quadratic form (often taught together with elementary linear algebra)? Unless I made a mistake, the asymptotes have slopes $(-5\pm\sqrt{41})/4$. – Jyrki Lahtonen Dec 15 '16 at 16:30
• Why do you think this has anything to do with number-theory? That is the study of prime numbers, divisibility, and the advanced topics that spawn from that? Did you read the tag description? – Jyrki Lahtonen Dec 15 '16 at 16:38
• @JyrkiLahtonen, Does theory of equation tag suits for this question ?? – Vidyanshu Mishra Dec 15 '16 at 16:40

What you have looks like:$$5x^2-5y^2-8xy-2x-4y+5=0$$

$$5^2+x(-8y-2)-5y^2-5y-5=0$$

$$5x^2-2x(4y+1)-y(5y+4)$$

See that it is a quadratic equation in terms of $x$. Solve for $x$ in terms of $y$.

Side Note: It represent a hyperbola and integer solutions are $(x,y)$=$(-5,10),(-1,2)$.

On rewritting the equation as

$-5y^2-y(8x+4)+5x^2-2x+5=0$

and solving it using quadratic formula we get

$y=\frac{(8x+4) \pm 2\sqrt{(-9x^2+26x-21}}{10}$

try to solve this for $y$ in terms of $x$

• This is wrong. The $x^2$ and $y^2$ terms should have opposite sign. There are solutions to this equation (see my comment on the original question). – MPW Dec 15 '16 at 16:29
• Still wrong. It should be "$-5y^2$" for the first term. – MPW Dec 15 '16 at 16:38