am i right?

Question

ok this is another quadratic thing...
x^2+y^2=2
y=x-2
so far this is what i got
x^2(x-2)(x-2)=2 x^2+x^2-2x-2x+4=2 2x^2-4x+4=2
-2 -2
2x^2-4x+2=0
any ideas?

Answer 1

you buggered up early. y doesn't equal x-2:
X^2 + y^2=2
rearrange for y by itself:
y^2 = 2-x^2
so :
y = squareroot(2-x^2)
this doesn't equal x-2
Try it with the squareroot answer and see how it goes.

Answer 2

you can then get rid of a factor of 2
x^2 -2x +1 = 0
which is (x -1)^2
so x = 1
and y = 1 - 2 = -1

Answer 3

2x^2-4x+2=0.
Divide by 2 to get x^2-2x+1=0
(x-1)^2 =0
x-1 = 0
x=1
y=-1
(1,-1) is (x,y) intersection

You can type in the functions at http://www.coolmath.com/graphit/index.html
to see the graphs.

There type in -sqrt(2-x^2) [Hit Eval]
Then type in sqrt(2-x^2) [Hit Eval]
Then type in x-2 [Hit Eval]
A graph of the intersecting functions will show on the right.

This site should help with any other graphing issues.

Answer 4

u said x^2+y^2=2 and y=x-2
I say yes because if u sustitute x with 1 u would get

(1x1)+(-1) x (-1)=2

1+1=2

Answer 5

after you got "lost", divide both sides by 2 leaves you with:

x^2-2x-1=0
(x-1)^2=0
x=1
therefore y=-1

Answer 6

Wow cool is this like a alien language cool are you from mars???

Answer 7

nope