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anonymous
 5 years ago
x  1 = x2 2x + 1
anonymous
 5 years ago
x  1 = x2 2x + 1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x1=x1^2 therefore the real solution is x=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to use the definition of absolute value and notice a couple of things. First,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[u:=\sqrt{u^2}\]and so\[\sqrt{(x1)^2}=\sqrt{(x^22x+1)^2}\]Square both sides and note you can factor the righthand side as such:\[(x1)^2=((x1)^2)^2\]Hence,\[(x1)^2=(x1)^4 \rightarrow (x1)^2(1(x1)^2)=0\]Therefore, either\[(x1)^2=0 \rightarrow x=1\]or\[1(x1)^2=0 \rightarrow (x1)^2=1 \rightarrow x1=\pm 1 \rightarrow x=1 \pm 1\]That is,\[x=0,2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So there are three solutions: x=0 x=1 x=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a plot of what you're trying to do. You have two functions,\[y=x1\]and\[y=x^22x+1\]and you're trying to find the points of intersection.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0THANK YOU SO SO MUCH LOKISAN AND IAMPAUL!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could you also help me with this , thanxs! 2x  1 = x2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I need a second to do something...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problemo but i guess lokisan explains it miles better ;)
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