4x^2-y^2+2y-1=0 ?? is two straight line at Cartesian level ... can anyone hepl me ?

Guest May 15, 2017

#1**+2 **

This is an interesting question :)

4x^2-y^2+2y-1=0 ?? is two straight line at Cartesian level ... can anyone hepl me ?

\(4x^2-y^2+2y-1=0 \\ 4x^2-(y^2-2y+1)=0 \\ 4x^2-(y-1)^2=0 \\ [2x-(y-1)][2x+(y-1)]=0\\ so\\ 2x-(y-1)=0 \qquad or \qquad 2x+(y-1)=0\\ 2x-y+1=0 \qquad \quad or \qquad 2x+y-1=0\\ y=2x+1 \qquad \qquad or \qquad y=-2x+1\\ \)

check:

Melody
May 15, 2017

#2**+2 **

**4x^2-y^2+2y-1=0 ?? is two straight line at Cartesian level**

\(\begin{array}{|rcll|} \hline 4x^2-y^2+2y-1 &=& 0 \\ 4x^2-(y^2-2y+1) &=& 0 \quad & | \quad y^2-2y+1 = (y-1)^2 \\ 4x^2-(y-1)^2 &=& 0 \\ (y-1)^2 &=& 4x^2 \quad & | \quad \text{square root both sides} \\ y-1 &=& \pm 2x \\ y &=& \pm 2x + 1 \\\\ \text{Line 1 }: y &=& 2x + 1 \\ \text{Line 2 }: y &=& -2x + 1 \\ \hline \end{array}\)

heureka
May 15, 2017