Question

help :) verify my answer :o

Answer 1

i got different value for y
\(\large y=+-\frac {\sqrt 7 }{2}\)

Answer 2

u can use the quadratic formula. u would have
\[\large z=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm\sqrt{-3}}{2}=
\frac{-1\pm i\sqrt{3}}{2} \]

Answer 3

this can be rewritten in ordered pair form as:
\[\large \left(\frac{-1}{2},\pm\frac{\sqrt{3}}{2}\right) \]

Answer 4

i know but its important to use the given method :O

Answer 5

OK

Answer 6

it would be like this
\(\large (x^2-y^2,2xy)+(x,y)+(1,0)=(0,0)\)
so i would have two equation
\(\large x^-y^2+x+1=0\)
\(\large 2xy+y=0 ---> y\neq 0 , x=\frac{-1}{2}\)

Answer 7

let z=(x,y) for brevity:
\[\large z^2+z+1=0 \]
\[\large (x,y)^2+(x,y)+(1,0)=(0,0) \]
\[\large (x^2-y^2,2xy)+(x,y)+(1,0)=(0,0) \]
\[\large (x^2-y^2+x+1,2xy+y)=(0,0) \]

Answer 8

Yes. u did it right!

Answer 9

\(\Huge y^2=( \frac{-1}{2})^2+\frac{-1}{2}+1\)
\(\Huge y=\sqrt\frac {1-2+3}{4}\)
oh lol nw i see i forget - sighn !!
thx alote !

Answer 10

since x=-1/2 then
\[\large (-1/2)^2-y^2+(-1/2)+1=0 \]
\[\large 1/4-1/2+1=y^2 \]
\[\large -1/4+1=y^2 \]
\[\large y^2=3/4 \]
\[ y=\pm\sqrt{3/4}=\pm\sqrt{3}/2 \]

Answer 11

u r welcome

Answer 12

thank you ^_^