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Math Help - Solve this equation (involves radicals)

  1. #1
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    Solve this equation (involves radicals)

    Hello all, Need your help

    Solve in [-1,1] the equation : \; (x+1)^2 = \sqrt{(1-x)^3}
    Last edited by Jhevon; April 11th 2009 at 04:54 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by linda2005 View Post
    Hello all, Need your help

    Solve in [-1,1] the equation : \; (x+1)^2 = \sqrt{(1-x)^3}
    Hint: square both sides and expand the brackets
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    Yes I do that , but x(x^3 + 5x^2 + 6x + 7)=0 it's so hard to resolve !
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    Quote Originally Posted by linda2005 View Post
    Hello all, Need your help

    Solve in {\color{red}[-1,1]} the equation : \; (x+1)^2 = \sqrt{(1-x)^3}
    Quote Originally Posted by linda2005 View Post
    Yes I do that , but x(x^3 + 5x^2 + {\color{red}3} x + 7)=0 it's so hard to resolve ! Mr F says: Note the red.
    The key phrase in the question is "Solve in [-1, 1] ....". It is clear that the (only) solution to the cubic lies outside this domain ....

    So the only solution is x = 0.
    Last edited by mr fantastic; April 11th 2009 at 05:16 PM. Reason: Added extra quote
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  5. #5
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    Quote Originally Posted by linda2005 View Post
    Yes I do that , but x(x^3 + 5x^2 + 6x + 7)=0 it's so hard to resolve !
    It's actually

    x(x^3 + 5x^2 + 3x + 7) = 0.

    Clearly x = 0 is a solution.

    Try graphing the equation x^3 + 5x^2 + 3x + 7.

    Clearly it has one root at about x = -4.65, but I don't believe there's a way to solve it exactly (I may be wrong).

    You might want to use a Heuristic method like the Bisection Method or Newton's Method to increase the accuracy.

    Edit: And as Mr F says, it doesn't lie in the domain anyway... this post is for future reference :P
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    for x \geq -1 we have: x^3 + 5x^2 + 3x + 7 \geq x^3 + 3x + 7 \geq -1 - 3 + 7 = 3. so x^3 + 5x<br />
^2 + 3x + 7 \neq 0, for all x \geq -1.
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    Quote Originally Posted by NonCommAlg View Post
    for x \geq -1 we have: x^3 + 5x^2 + 3x + 7 \geq x^3 + 3x + 7 \geq -1 - 3 + 7 = 3. so x^3 + 5x<br />
^2 + 3x + 7 \neq 0, for all x \geq -1.
    Like I said ... "It is clear ....."
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  8. #8
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    Quote Originally Posted by linda2005 View Post

    Hello all, Need your help

    Solve in [-1,1] the equation : \; (x+1)^2 = \sqrt{(1-x)^3}
    was x \geq -1 a part of problem or ...? because the only restriction that we need here is x \leq 1 to make sure that the right hand side of the equation is defined.
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