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Math Help - Easiest way to solve

  1. #1
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    Easiest way to solve

    What is the simplest way to solve this problem?

    |x+1|^2+3|x+1|-4=0
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  2. #2
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    Hello, Mike9182!

    What is the simplest way to solve this equation?

    . . |x+1|^2+3|x+1|-4\:=\:0
    We have a quadratic . . . Solve it accordingly.


    Factor: . \left(|x+1| + 4\right)\left(|x+1| - 1\right) \:=\:0


    Set each factor equal to zero and solve:

    . . |x+1| + 4 \:=\:0 \quad\Rightarrow\quad|x+1| \:=\:-4 . . . no roots
    . . . .
    An absolute value is never negative.

    . . |x+1| - 1 \:=\:0 \quad\Rightarrow\quad |x+1| \:=\:1 \quad\Rightarrow\quad x+1 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:0,\:2


    But x=2 is extraneous.

    . . Therefore, the only solution is: . x \:=\:0

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  3. #3
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    Absolute value

    Hello Mike9182
    Quote Originally Posted by Mike9182 View Post
    What is the simplest way to solve this problem?

    |x+1|^2+3|x+1|-4=0
    Let y = |x+1|. Then:

    y^2 +3y - 4=0

    \Rightarrow (y+4)(y-1)=0

    \Rightarrow y=-4, \,1

    But y = -4 is impossible, since |x+1| \ge 0

    \Rightarrow y =1= |x+1|

    \Rightarrow x+1 = \pm 1

    \Rightarrow x = 0,\, -2

    Grandad
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Soroban View Post
    x+1 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:0,\:2
    Am I missing something?
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  5. #5
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    Quote Originally Posted by alexmahone View Post
    Am I missing something?
    probably just a typo ... it happens.
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