Question

Find real points satisfying 3x^2+3y^2-4xy+10x-10y+10=0

Answer 1

Given
$3x^2+3y^2-4xy+10x-10y+10=0$
This can be rewritten as follows:
$3y^2+(-4x-10)y+(10+10x+3x^2)=0$

Notice that this is a quadratic equation of the form:
$ay^2+by+c=0$,
with the solution given by:
$y= \frac{-b\pm \sqrt{b^2-4ac} }{2a}$
where:
$a = 3, \\ b = -4x - 10 \\ c = 10 + 10x + 3x^2$

Thus,
$y= \frac{-(-4x-10)\pm \sqrt{(-4x-10)^2-4(3)(10+10x+3x^2)} }{2(3)} \\ \\ =\frac{4x+10\pm \sqrt{(16x^2+80x+100)-12(10+10x+3x^2)} }{6} \\ \\ =\frac{4x+10\pm \sqrt{16x^2+80x+100-120-120x-36x^2} }{6} \\ \\ =\frac{4x+10\pm \sqrt{-20x^2-40x-20} }{6} \\ \\ =\frac{4x+10\pm 2\sqrt{-5x^2-10x-5} }{6}$

Now consider
$\sqrt{-5x^2-10x-5}$
This gives a real solution if and only if
$-5x^2-10x-5 \geq 0$
i.e.
$-5(x^2+2x+1) =0 \\ \\ (x+1)^2=0 \\ \\ x+1=0 \\ \\ x=-1$

Thus,
$-5x^2-10x-5 \geq 0$
when x = -1

Thus, the equation has real solutions only when x = -1 at which the value of y is given by
$y=\frac{4(-1)+10\pm 2\sqrt{-5(-1)^2-10(-1)-5} }{6} \\ \\ =\frac{-4+10\pm 2\sqrt{-5+10-5} }{6} =\frac{6\pm 2\sqrt{0} }{6} =\frac{6\pm 2(0) }{6} \\ \\ =\frac{6\pm 0 }{6} = \frac{6}{6} =1$

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