Question

Fnd the real points $(x,y)$ satisfying $3x^2+3y^2-4xy+10x+10y+10=0.$

Answer 1

Consider the following equation.

$3x^2+3y^2-4xy+10x-10y+10=0$

This can be rewritten as,

$3y^2+(-4x-10)y+(10+10x+3x^2)=0$

Notice that this is the quadratic equation of the form $a{y}^2+by+c=0$.

Using discriminant formula, we have

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where, $a=3$, $b=-4x-10$, $c=10+10x+3x^2$

$y=\frac{\{-(-4x-10)\pm \sqrt{{(-4x-10)}^2-4\times 3(10+10x+3x^2)}\}}{2\times 3}$

$y=\frac{\{4x+10\pm \sqrt{(16x^2+80x+100)-12(10+10x+3x^2)}\}}{6}$

$y=\frac{4x+10\pm \sqrt{16x^2+80x+100-120-120x-36x^2}}{6}$

$y=\frac{4x+10\pm \sqrt{-20x^2-40x-20}}{6}$

$y=\frac{4x+10\pm 2\sqrt{-5x^2-10x-5}}{6}$

Now, consider $\sqrt{-5x^2-10x-5}$.

This gives real solutions if and only if,

$-5x^2-10x-5\ge 0$

$-5(x^2+2x+1)=0$

${(x+1)}^2=0$

$x=-1$

Thus,

$-5x^2-10x-5\ge 0$

$where,$

$x=-1$

Thus, the equation has real solutions only when $x=-1$ at which the value of $y$ will be,

$y=\frac{4(-1)+10\pm 2{\sqrt{-5{(-1)}^2-10(-1)-5}}^{}}{6}$

$y=\frac{-4+10\pm 2\sqrt{-5+10-5}}{6}$

$y=\frac{6\pm 2\sqrt{0}}{6}$

$y=1$

Therefore, the real point satisfying $3x^2+3y^2-4xy+10x-10y+10=0$ is $x=-1$ and $y=1$.