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Thread: Easiest way to solve

  1. #1
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    Easiest way to solve

    What is the simplest way to solve this problem?

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

    What is the simplest way to solve this equation?

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


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


    Set each factor equal to zero and solve:

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

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


    But $\displaystyle x=2$ is extraneous.

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

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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?

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

    $\displaystyle y^2 +3y - 4=0$

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

    $\displaystyle \Rightarrow y=-4, \,1$

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

    $\displaystyle \Rightarrow y =1= |x+1|$

    $\displaystyle \Rightarrow x+1 = \pm 1$

    $\displaystyle \Rightarrow x = 0,\, -2$

    Grandad
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Soroban View Post
    $\displaystyle x+1 \:=\:\pm1 \quad\Rightarrow\quad x \:=\:0,\:2$
    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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