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Math Help - Solving Quadratic Systems

  1. #1
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    Solving Quadratic Systems

    Find the exact solutions of x^2 - (y - 5)^2 = 25 and y = -x^2


    Do I try to isolate the y in the first equation first?
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    Quote Originally Posted by BlueStar View Post
    Find the exact solutions of x^2 - (y - 5)^2 = 25 and y = -x^2


    Do I try to isolate the y in the first equation first?
    you could, but i wouldn't. replace the y in the first equation with the -x^2 and then solve for x

    once you have x, plug it back into the y = -x^2 equation to solve for y
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  3. #3
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    Would that be like x^2 - (-x^2 - 5)^2 = 25? Or should I just drop the exponent since one is over the parenthesis?
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    Quote Originally Posted by BlueStar View Post
    Would that be like x^2 - (-x^2 - 5)^2 = 25? Or should I just drop the exponent since one is over the parenthesis?
    no, that's correct. solve that equation for x
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  5. #5
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    Is the answer 0?


    x^2 - (-x^2 - 5)^2 = 25
    x^2 + (x^2 + 5)^2 = 25
    x^2 + x^2 + 5 = 5
    x^2 + x^2 = 0
    2x^2 = 0
    x^2 = 0
    x = 0
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    Quote Originally Posted by BlueStar View Post
    Is the answer 0?


    x^2 - (-x^2 - 5)^2 = 25
    x^2 + (x^2 + 5)^2 = 25
    x^2 + x^2 + 5 = 5
    x^2 + x^2 = 0
    2x^2 = 0
    x^2 = 0
    x = 0
    remember how to expand binomial squares: (a \pm b)^2 = a^2 \pm 2ab + b^2

    here, you a is -x^2 and your b is -5
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    So would the solutions be (0, -5)?
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    Quote Originally Posted by BlueStar View Post
    So would the solutions be (0, -5)?
    x = 0 is NOT a solution (if we have x = 0, that would lead to -25 = 25), and if it was, since y = -x^2, if x = 0, then y = 0, so a solution would be (0,0)

    I'll start you off.

    x^2 - \left( -x^2 - 5 \right)^2 = 25

    \Rightarrow x^2 - \left( x^4 + 10x^2 + 25 \right) = 25

    \Rightarrow x^2 - x^4 - 10x^2 - 25 = 25

    Now continue (it seems as though the solutions are complex, as in, not real)
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    -x^4 -9x^2 = 50


    ??
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    Quote Originally Posted by BlueStar View Post


    -x^4 -9x^2 = 50


    ??
    Note that we can rewrite this as: x^4 + 9x^2 + 50 = 0

    which in turn we could rewrite as: \left( x^2 \right)^2 + 9 \left( x^2 \right) + 50 = 0

    which is a quadratic eqaution in x^2, we can therefore employ the quadratic formula
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  11. #11
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    x^2 + 9x + 50 = 0

    x = -9 +- /(9)^2 - 4(1)(50) (attempt at typing quadratic formula)
    2(1)
    x = -9 +- /81 - 200
    2


    But the square root number turns out to be negative...
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  12. #12
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    Quote Originally Posted by BlueStar View Post
    x^2 + 9x + 50 = 0

    x = -9 +- /(9)^2 - 4(1)(50) (attempt at typing quadratic formula)
    2(1)
    x = -9 +- /81 - 200
    2


    But the square root number turns out to be negative...
    yes, that's supposed to happen, as i said, this equation has complex solutions. have you done solutions like that in class?
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    Now I'm vaguely remembering an imaginary term, i, fitting into the equation. So would it be -119i, or something like that?
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    Quote Originally Posted by BlueStar View Post
    Now I'm vaguely remembering an imaginary term, i, fitting into the equation. So would it be -119i, or something like that?
    no, it would be x^2 = \frac {-9 \pm \sqrt {119}~i}{2}

    so now, you can solve that value for x, but you may want to plug in the expression for x^2 into y = -x^2 first
    Last edited by Jhevon; July 23rd 2007 at 05:50 PM.
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  15. #15
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    You mean y = 9 +- 119i / -2?
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