Question

solve xp² +(y-x)P-y=0​

Answer 1

Answer:

xp^2+(y-x)p-y=0

=> xp^2+py-px-y=0

=> p(xp+y)-1(xp+y)=0

=> (xp+y)(p-1)=0

=> xp+y=0

=> xp= -y

=> p= -y/x

p-1=0

=> p=1

Answer 2

The given equation is $xp^2+(y-x)p-y=0$

According, to the question.

$xp^2+(y-x)p-y=0$

$xp^2+yp-xp-y=0$

$xp^2-xp+yp-y=0$

$xp(P-1)+y(P-1)=0$

$(P-1)(xp+y)=0$

On compare,

$p=1, p=-\frac{4}{x}.$