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Math Help - Weird Basic Algebra Problem (Help Appreciated =D )

  1. #1
    FIRESSS
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    Question Weird Basic Algebra Problem (Help Appreciated =D )

    Hello,

    I got this annoying problem I can't seem to get around, I know there's a trick somewhere in the book for this...I think there is, anyhow, I need to solve the following equation for x:

    (s^2 - 2sx + x^2)(a) = (x^2)(b)

    Each time I try to mess around with x and s, I'm unable to break the 2sx term and it's rly getting to me, at my last step I got to this useless mess:

    (s^2 - 2sx)/(x^2) = (b/a)(x^2)

    Any tips/tricks highly appreciated, spanks!
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by FIRESSS View Post
    Hello,

    I got this annoying problem I can't seem to get around, I know there's a trick somewhere in the book for this...I think there is, anyhow, I need to solve the following equation for x:

    (s^2 - 2sx + x^2)(a) = (x^2)(b)

    Each time I try to mess around with x and s, I'm unable to break the 2sx term and it's rly getting to me, at my last step I got to this useless mess:

    (s^2 - 2sx)/(x^2) = (b/a)(x^2)

    Any tips/tricks highly appreciated, spanks!
    Hello,

    if you factorize the first bracket you'll notice that it is a complete square: s^2-2sx+x^2=(s-x)^2

    After a few steps you'll get the proportion:

    \frac{(s-x)^2}{x^2}=\frac{a}{b} \Longleftrightarrow \left( \frac{s-x}{x} \right)^2=\frac{a}{b}

    EB
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  3. #3
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    Hello, FIRESSS!

    Solve for x:\;\;a(s^2-2sx+x^2)\;=\;bx^2

    Why not expand it and solve the resulting quadratic equation?


    We have: . as^2 -2asx + ax^2\;=\;bx^2\quad\Rightarrow\quad(a-b)x^2 - 2asx + as^2\;=\;0


    Quadratic Formula: . x \;=\;\frac{-(\text{-}2as) \pm \sqrt{(\text{-}2as)^2 - 4(a-b)(as^2)}}{2(a-b)}

    . . which simplifies to: . x \;=\;\frac{as \pm s\sqrt{ab}}{a-b}

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