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Thread: Quadratic Equations

  1. #1
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    Quadratic Equations

    Need help in solving the following quadratic equations to the power of three...

    $\displaystyle x^3-y^3=35$
    $\displaystyle x-y=5$

    Pls explain or show any relevant working. Thanks in advance for your help.
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  2. #2
    A Plied Mathematician
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    What have you tried thus far?
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  3. #3
    Senior Member eumyang's Avatar
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    Use substitution. Solve the 2nd equation for x:
    $\displaystyle x = y + 5$

    Substitute this into the 1st equation for x:
    $\displaystyle \begin{aligned}
    x^3 - y^3 &= 35 \\
    (y + 5)^3 - y^3 &= 35 \\
    y^3 + 15y^2 + 75y + 125 - y^3 &= 35 \\
    15y^2 + 75y + 125 &= 35 \\
    15y^2 + 75y + 90 &= 0
    \end{aligned}$
    Solve this quadratic (you'll get two answers for y). Plug both into the equation
    $\displaystyle x = y + 5$
    to find the corresponding x values.
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