Question

The number of distinct pairs $(x,y)$ of real numbers that satisfy the equation
$4x^2+4xy+2y^2-2y+1=0$ is

• A. $0$
• B. $1$
• C. $4$
• D. infinitely many

Answer 1

Answer: (B)

$1$

$4x^2+4xy+2y^2-2y+1=0$
$\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$=4\left(2\right)\left(1\right)+2\left(0\right)(-1)\left(2\right)-4(-1{)}^2-2(0{)}^2-1(2{)}^2$
$=8-4-4$
$=0$
$\therefore$it represents a pair of lines which intersect at
$(\frac{\partial s}{\partial x}=0,\frac{\partial s}{\partial y}=0)$
$\Rightarrow 8x+4y=0;4x+4y-2=0$
$\Rightarrow y=-2x$
$\Rightarrow 4x+4(-2x)=2$
$\Rightarrow -4x=2$
$\Rightarrow x=\frac{-1}{2}$
$\Rightarrow y=1$
$\therefore$ only one point.