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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    |x-1|=|x-1|^2 therefore the real solution is x=1

  2. anonymous
    • 5 years ago
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    You need to use the definition of absolute value and notice a couple of things. First,

  3. anonymous
    • 5 years ago
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    \[|u|:=\sqrt{u^2}\]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
    • 5 years ago
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    So there are three solutions: x=0 x=1 x=2

  5. anonymous
    • 5 years ago
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  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    THANK YOU SO SO MUCH LOKISAN AND IAMPAUL!

  8. anonymous
    • 5 years ago
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    np :)

  9. anonymous
    • 5 years ago
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    could you also help me with this , thanxs! |2x - 1| = x2

  10. anonymous
    • 5 years ago
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    ok

  11. anonymous
    • 5 years ago
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    I need a second to do something...

  12. anonymous
    • 5 years ago
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    no problemo but i guess lokisan explains it miles better ;)

  13. anonymous
    • 5 years ago
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    :)

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