Question

If (x,y) be a real solution of the equation $x^2+12xy+4y^2+4x+8Y+20=0$ then prove that $x\notin (-2,1)andy\notin [-1,\frac{1}{2}]$

Answer 1

$x^2+12xy+4y^2+8y+4x+20=0$

Partially differentiate w.r.t. x by

$2x+12y+0+0+4=0$

$x+6y+2=0$ ...(1)

$0+12x+8y+8+0+0=0$

$3x+2y+2=0$ ...(2)

$3\times \left(1\right)$

$3x+18y+6=0$

$3x+2y+2=0$

_______________

- - -

$16y+4=0$

$y=-\frac{1}{4}$

$x-\frac{3}{2}+2=0$

$x=-2+\frac{3}{2}$

$x=\frac{-1}{3}$

So, we can say that $x\notin (-2,1)$ & $y\notin [-1,\frac{1}{2}]$

Hence proved