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Mathematics 11 Online
OpenStudy (anonymous):

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

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

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

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

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.

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

OpenStudy (anonymous):

plus i already have the answers

OpenStudy (anonymous):

is the annswer x=0 and x=1

myininaya (myininaya):

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