## anonymous 5 years ago |x - 1| = |x2 -2x + 1|

1. anonymous

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

2. anonymous

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

3. anonymous

$|u|:=\sqrt{u^2}$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,$x=0,2$

4. anonymous

So there are three solutions: x=0 x=1 x=2

5. anonymous

6. anonymous

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.

7. anonymous

THANK YOU SO SO MUCH LOKISAN AND IAMPAUL!

8. anonymous

np :)

9. anonymous

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

10. anonymous

ok

11. anonymous

I need a second to do something...

12. anonymous

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

13. anonymous

:)