# all real numbers $y$,$x$ such that :$8x^2-2xy^2=6y=3x^2+3x^3y^2$

How to find the all real numbers $y$,$x$ such that :

$$8x^2-2xy^2=6y=3x^2+3x^3y^2$$

-

## 1 Answer

Use the resultant of $8x^2-2xy^2-6y, 3x^2+3x^3y^2-6y$ with respect to the variable $y$. This gives a polynomial that is satisfied by the $x$-coordinate of any of the common solutions. In this case the resultant factors as $36x^3(x-1)(16x^6+16x^5+24x^4+24x^3+25x^2+10x+10)$. This has the real roots $x=0, x=1$ only. So we get the solutions $(0,0) (1,1)$.

-
Can you explain ,how did you find $36x^3(x-1)(16x^6+16x^5+24x^4+24x^3+25x^2+10x+10)$ ? –  Frank Aug 9 '12 at 4:30
To compute the resultant see en.wikipedia.org/wiki/Sylvester_matrix –  i. m. soloveichik Aug 11 '12 at 0:25