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Math Help - Literal Equations

  1. #1
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    Unhappy Literal Equations

    Help, please.
    I understand the basics of solving literal equations. I can solve basic equations like P = srt.

    But I am stumped on these:

    x + SQRT(x) = -y
    Solve for x

    l/w = w/(l w)
    Solve for w

    Can anyone help??
    Thanks for any assistance anyone can provide!!!
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  2. #2
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    Quote Originally Posted by n2hdcycles
    Help, please.
    I understand the basics of solving literal equations. I can solve basic equations like P = srt.

    But I am stumped on these:

    x + SQRT(x) = -y
    Solve for x

    l/w = w/(l w)
    Solve for w

    Can anyone help??
    Thanks for any assistance anyone can provide!!!
    LITERAL?!? You mean linear equations.
    Well, these equations are called quadratic equations, meaning the x vaiable is now squared. Have you studied those?
    I can solve them for you but if you never studied quadradic equations then it would be useless to show you how to solve them.
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  3. #3
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    literal (??) equation

    Yes I understand the quadratic formula. Any help in guiding me through the solutions would be appreciated. I am really stumped on these two.

    It is my understanding that a literal equation is a linear equation that is all letters (no numbers). "Literal" means "related to or being comprised of letters". See the following link:
    http://www.purplemath.com/modules/solvelit.htm

    Like I said, I can get basic equations, like P = 2l + 2w solved for l.
    P = 2l + 2w
    p - 2w = 2l
    (p - 2w)/2 = l

    It is the square root on the first equation and the fraction on the second equation that's throwing me.

    Again -- appreciate the assistance!!
    Last edited by n2hdcycles; January 8th 2006 at 03:30 PM.
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  4. #4
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    Do you know the quadratic formula? If you do I can guide you through those two.
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  5. #5
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    Yes, I do understand the quadratic formula.
    Sorry, after I posted a reply. I realized I neglected to answer that question.
    I've spent many hours attempting to work these problems out.... very frustrated......mmmmmmmmmm...and slightly grouchy for a sunday night
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  6. #6
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    For the first problem
    x+\sqrt{x}=-y
    You have to do the following trick, let \sqrt{x}=z thus, x=z^2. Substitute that back to the original equation:
    z^2+z=-y
    z^2+z+y=0
    Use quadratic formula,
    z=\frac{-1\pm \sqrt{1-4y}}{2}
    But, z=\sqrt{x}
    \sqrt{x}=\frac{-1\pm \sqrt{1-4y}}{2}
    Now square both sides (do one for plus and then for minus):
    Remember to use F.O.I.L.

    x=[\frac{-1+\sqrt{1-4y}}{2}]^2=\frac{2-4y+2\sqrt{1-4y}}{4}= 1/2-y+\frac{\sqrt{1-4y}}{2}

    Now do the same thing for the minus,

    x=[\frac{-1-\sqrt{1-4y}}{2}]^2=\frac{2-4y-2\sqrt{1-4y}}{4}= 1/2-y-\frac{\sqrt{1-4y}}{2}

    I would say if you are starting to learn algebraic equations this equation was very difficult
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  7. #7
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    For the second,
    \frac{l}{w}=\frac{w}{l-w} for l
    Cross multiply,
    l(l-w)=w^2
    Open parantheses,
    l^2-wl=w^2
    Organize to quadratic form,
    l^2-wl-w^2=0
    Apply Qudratic formula,
    l=\frac{w\pm \sqrt{w^2+4w^2}}{2}
    Add them up,
    l=\frac{w\pm \sqrt{5w^2}}{2}
    But, \sqrt{5w^2}=w\sqrt{5}
    Thus,
    l=\frac{w\pm w\sqrt{5}}{2}
    Factor w
    l=w[1\pm \frac{\sqrt{5}}{2}]

    It is interesting to note this problem was about the Golden Mean which appears in this solution represented by \phi it is the positive solution for l thus, \phi=\frac{1+\sqrt{5}}{2}. It makes many interesting appearances. If you are curious about the Golden Mean go here, http://en.wikipedia.org/wiki/Golden_ratio this is a truly remarkable explanation.
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  8. #8
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    digesting your information

    I followed the explanation for equation #1. I understand it perfectly. THANK YOU!

    Equation #2.
    I need to solve for w (not l). But I should be able to get there by the example you have given me. I will give it a shot.
    Thanks for the website link -- the question is further in depth and I need to understand the "golden rectangle". So i'm sure it will come in handy!

    You have been a tremendous help!
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