Question

|x - 1| = |x2 -2x + 1|

Answer 1

|x-1|=|x-1|^2
therefore the real solution is x=1

Answer 2

You need to use the definition of absolute value and notice a couple of things. First,

Answer 3

\[|u|:=\sqrt{u^2}\]and so\[\sqrt{(x-1)^2}=\sqrt{(x^2-2x+1)^2}\]Square both sides and note you can factor the right-hand side as such:\[(x-1)^2=((x-1)^2)^2\]Hence,\[(x-1)^2=(x-1)^4 \rightarrow (x-1)^2(1-(x-1)^2)=0\]Therefore, either\[(x-1)^2=0 \rightarrow x=1\]or\[1-(x-1)^2=0 \rightarrow (x-1)^2=1 \rightarrow x-1=\pm 1 \rightarrow x=1 \pm 1\]That is,\[x=0,2\]

Answer 4

So there are three solutions:

x=0
x=1
x=2

Answer 5

This is a plot of what you're trying to do. You have two functions,\[y=|x-1|\]and\[y=|x^2-2x+1|\]and you're trying to find the points of intersection.

Answer 6

THANK YOU SO SO MUCH LOKISAN AND IAMPAUL!

Answer 7

np :)

Answer 8

could you also help me with this , thanxs! |2x - 1| = x2

Answer 9

ok

Answer 10

I need a second to do something...

Answer 11

no problemo but i guess lokisan explains it miles better ;)

Answer 12

:)