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Math Help - Radical Equation

  1. #1
    VAP
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    Radical Equation

    I could not make any headway on this problem. Thanks for viewing and helping.

    Using:
    4 √x = 2x + k
    find three different expressions to substitute for k so that the equation has two, one, and no solutions. Describe how you found the equations.
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  2. #2
    A riddle wrapped in an enigma
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    CORRECTED VERSION

    1. Square both sides of the equation to get:


    2. Set the equation = 0


    3. Factor x from the two middle terms:


    4. Find the discriminant. From the quadratic formula, the discriminant is what's under the radical, namely:


    You can determine the nature of the roots by inspecting the discriminant (D). If D=0, then you have one real root; if D>0, then you have 2 real roots, and if D<0, you have no real roots.

    I found the discriminant to be:



    5. Simplifying:




    Made an error.....correcting here

    If D=0, then k=2, meaning you have one real root.




    If D>0, then 0<k<2, meaning you have two real roots.

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  3. #3
    Moo
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    Quote Originally Posted by VAP View Post
    I could not make any headway on this problem. Thanks for viewing and helping.

    Using:
    4 √x = 2x + k
    find three different expressions to substitute for k so that the equation has two, one, and no solutions. Describe how you found the equations.
    4 \sqrt{x}=2x+k

    Note that x=(\sqrt{x})^2

    So you can substitute y=\sqrt{x}.


    The equation is now 4y=2y^2+k

    Divide by 2, to try to get a known form of a quadratic :

    y^2-2y+\frac k2=0

    Completing the square, it will give :

    y^2-2y+1+\left(\frac k2-1\right)=0

    (y-1)^2=1-\frac k2

    It has only one solution if (y-1)^2=0... So k=\dots

    It has no solution if the right hand side is negative (because it's not possible that a square is negative). So k=\dots, for example

    It's ok if the right hand side is positive, but be careful, because y has to be positive (since \sqrt{x}>0). If I thought correctly, taking k>0 and such that 1-k/2>0, it will be ok



    I tried to make it without "discriminants" because I don't know if you have learnt this So it's not necessarily accurate, especially with the last part, for finding 2 solutions.
    Last edited by Moo; May 24th 2008 at 10:06 AM. Reason: extra info
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