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Math Help - Ratio...

  1. #1
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    Ratio...

    If 2x^2 - 3xy + y^2 = 0, find the ratio of x:y.
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  2. #2
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    Quote Originally Posted by snigdha View Post
    If 2x^2 - 3xy + y^2 = 0, find the ratio of x:y.
    The first step would be to divide through by y:

    \frac{2x^2}{y} - 3x + y = 0

    Then divide through by x:

    \frac{2x}{y} - 3 + \frac{y}{x} = 0

    Now, we move the last two terms to the other side of the equation:

    \frac{2x}{y} = 3 - \frac{y}{x}

    Then we divide through by 2:

    \frac{x}{y} = \frac{3}{2} - \frac{y}{2x}

    And there you go.
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  3. #3
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    Another way: divide through immediately by y^2.

    That gives you 2(x/y)^2- 3(x/y)+ 1= 0 a quadratic equation for x/y.
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  4. #4
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    Quote Originally Posted by snigdha View Post
    If 2x^2 - 3xy + y^2 = 0, find the ratio of x:y.
    Divide through by y^2

    2(x/y)^2 - 3(x/y) + 1 = 0

    Solve for (x/y). Factoring works:

    [2(x/y) - 1][(x/y) - 1] = 0

    (x/y) = 1/2 or 1 ...x:y = 1:2 or 1:1
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  5. #5
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    Hi snighda,
    the previous two methods given by HallsofIvy and Diagonal
    are the main variations.

    2x^2-3xy+y^2=0 may be converted to a quadratic in \frac{x}{y}

    \frac{2x^2-3xy+y^2}{y^2}=\frac{0}{y^2}

    2\left(\frac{x}{y}\right)^2-3\left(\frac{x}{y}\right)+1=0

    This is now of the form aq^2+bq+c=0

    for which the solutions are q=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Hence 2\left(\frac{x}{y}\right)^2-3\left(\frac{x}{y}\right)+1=0 for \frac{x}{y}=\frac{3\pm\sqrt{9-4(2)}}{2(2)}=\frac{3\pm1}{4}

    You achieve the exact same result by writing the expression as a multiplication of factors

    2x^2-3xy+y^2=(2x+u)(x+v)

    uv=y^2,\ 2xv+ux=x(2u+v)=x(-3y)

    therefore u=v=-y

    2x^2-3xy+y^2=0\ \Rightarrow\ (2x-y)(x-y)=0

    Two numbers multiplied give zero, hence either one of them can be zero.

    which yields two solutions, x=y and 2x=y

    from which the exact same ratios obtained from the quadratic may be written.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Another way: divide through immediately by y^2.

    That gives you 2(x/y)^2- 3(x/y)+ 1= 0 a quadratic equation for x/y.
    Sorry, Halls. I'm certain that yours was not there when I posted my solution. I will have to check my time zone or whatever more closely.
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