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Math Help - Constructing a Quadratic equation.

  1. #1
    Junior Member BobBali's Avatar
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    Constructing a Quadratic equation.

    The question requires me to construct an equation and solve for y.

    Q. When a mother's age is y^2 then the daughter's age is y. When the mother lives to the age of 13y then the daughter will be the same age of
     <br />
y^2<br />
    Find the age of the daughter?

    Ok here goes:  y^2 = y
    The mothers age >
    (y^2 + 13y)^2 = y^2
    (y^2 + 13y) (y^2 + 13y) = y^2
    y^4 +13y^3 + 13y^3 + 169y^2 = y^2
     y^4 + 26y^3 + 168y^2 = 0

     <br />
y^2(y^2 + 26y + 168) = 0<br /> <br />

    Now solve using quadratic formula a=1, b=26, c=168



    [tex] \frac{-26}{2} + - \frac{\sqrt{676-672}{2} [\MATH]



    [tex] -13 + - \frac{\sqrt{4}{2} > x= -12 or x=-14 ?
    [\MATH]

    So, daughter's age is 12 or 14? But that doesn't make sense because if the daughter's age at the start was 12, then mother's age would be 144! Not feasible..
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  2. #2
    MHF Contributor
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    I don't understand how you came up with the equations.

    Quote Originally Posted by BobBali View Post
    Ok here goes:  y^2 = y
    Where does this come from? Is this supposed to mean that y = 1?
    The mothers age >
    (y^2 + 13y)^2 = y^2
    Why do you add two mother's ages: the current one and the future one? And why do you square the result?

    Mother's current age is y^2. After some number of years, she is 13y.
    Daugter's current age is y. After the same number of years, she is y^2.

    From this, it seems clear how to construct an equation.
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  3. #3
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    Or...

    the mother and daughter's age difference is constant

    y^2-y=13y-y^2
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  4. #4
    Junior Member BobBali's Avatar
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    Yes Sir, thank you! A hint was later given to us the next day that we should try and keep the ages of the mother and daughter constant; which apparently you already saw... Thank you once again. Take care.

    Regards,
    Bob
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  5. #5
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    It's not that you need to somehow keep their ages constant,
    but for example, if you were 15 and your friend is 13,
    and you both have the same birthday...

    then when you are 16, your friend will be 14,
    when you are 17, your friend will be 15,
    when you are 18, your friend will be 16,
    when you were 5, your friend was 3.

    You will always be 2 years older than your friend.
    So the difference between the mother's age and her daughter's age is always the same
    (even if the difference would involve fractions or decimals, which doesn't happen in the question).
    Last edited by Archie Meade; February 18th 2011 at 05:12 PM.
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