find the points of intersection y=x⁴-2x+1, y=1-x^2

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OpenStudy (anonymous):

this is what i have so far
x⁴-2x+1=1-x^2
x⁴-2x+x^2=0

OpenStudy (anonymous):

if i take out an x im left with \[x(x ^{3}+x ^{2}-2)=0\]

OpenStudy (anonymous):

Two points of intersection.

OpenStudy (anonymous):

there's three points

myininaya (myininaya):

x(x^3+x-2)=0

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OpenStudy (anonymous):

(0,1)
and
(1,0)
and
(0,0)

myininaya (myininaya):

x=0 or x^3+x-2=0
well if x=1 we see that works
1 | 1 0 1 -2
| 1 1 2
--------------------
1 1 2 | 0
so we have x^2+x+2=0
we can use quadratic formula here

OpenStudy (anonymous):

okay i'm good now. i didnt think about that. thanks

OpenStudy (anonymous):

i think but keep going please

myininaya (myininaya):

\[x=\frac{-1 \pm \sqrt{1-4(1)(2)}}2\]
but that is not real

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myininaya (myininaya):

so we only have x=0 and x=1

OpenStudy (anonymous):

why not just grapgh it?

OpenStudy (anonymous):

what happened to your x^3

OpenStudy (anonymous):

need to show work bass

myininaya (myininaya):

if x=0 then y=1
if x=1 then y=0

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myininaya (myininaya):

remember i divided

myininaya (myininaya):

ok i will do it on paper k?

OpenStudy (anonymous):

what happened to you x^3

myininaya (myininaya):

i divided

OpenStudy (anonymous):

Wow, just neglect my answer. I graphed it and got the correct answers. If you don't believe it, its your loss mate.

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OpenStudy (anonymous):

but it was originally x^4

OpenStudy (anonymous):

abe i cant just graph it thats useless. theres no way for me to show work on a calculator