Solve quadratic equation in two variables .

Question

What should be the best approach to solve these types of equations.

$$3x^2-63+12y=0$$

$$3y^2-63+12x=0$$ Actually they are partial derivatives of this equation $f(x,y)=x^3+y^3-63(x+y)+12xy$

I am doing maxima minima questions of two variables and i am really having hard time solving these equations.

Answer 1

By equating we obtain

$$3x^2-3y^2=12y-12x\iff3(x-y)(x+y)=12(y-x)\iff y=x \quad \lor \quad x+y=4$$

and thus

• for $y=x$

$$3x^2-63+12x=0\implies x=-7,3 \implies (x,y)=(-7,-7) \quad (x,y)=(3,3)$$

• for $y=4-x$

$$3x^2-63+12(4-x)=0\implies x=-1,5 \implies (x,y)=(-1,5) \quad (x,y)=(5,-1)$$

Answer 2

Note that on subtracting the two equations, we get: $$3(x^2-y^2)\color{red} -12(x-y)=0$$ $$\implies (x-y)\left[3(x+y)\right]-12(x-y)=0$$ $$\implies (x-y)(3x+3y-12)=0$$

What can you infer from this?

Answer 3

You can easily solve this without any use of derivatives.

$y^2-21+4x=0$

$y^2=21-4x$

$y=\sqrt{21-4x}$

$3x^2-63+12\sqrt{21-4x}=0$

$x^2-21=-4\sqrt{21-4x}$

$x^4-42x^2+441=16(21-4x)$

Can you solve it from here?