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Math Help - Quadratic Equation help

  1. #1
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    Quadratic Equation help

    For some reason my calculator wont seem to work this out...help please



    0 = 5.887E24x^2 - 4.648E33x + 9.059E41



    The "E"s are meant to represent scientific notation. You could replace them with a "5.887x10^5" type of thing, but I thought that would have made it more confusing.

    Yes, I have tried the quadratic equation (the negative "b" plus or minus the square root of "b" squared minus BLAH BLAH BLAH). I know the equation by heart, and tried it in my calculator. Could someone please help with this.
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  2. #2
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    Quote Originally Posted by zach610 View Post

    Yes, I have tried the quadratic equation (the negative "b" plus or minus the square root of "b" squared minus BLAH BLAH BLAH). I know the equation by heart, and tried it in my calculator. Could someone please help with this.
    Then you should have the answer
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Then you should have the answer
    I tried that, but it gives me the same answer with the plus, and the minus. When, instead it is supposed to give you one answer negative, and the other is positive.
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  4. #4
    Super Member bigwave's Avatar
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    what calculator are you using

    what calculator are you using

    TI89 ???
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  5. #5
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    Quote Originally Posted by bigwave View Post
    what calculator are you using

    TI89 ???
    TI-84 Plus

    I got one of the old ones, without the equation solver
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by zach610 View Post
    I tried that, but it gives me the same answer with the plus, and the minus. When, instead it is supposed to give you one answer negative, and the other is positive.
    Not always. If b^2-4ac=0 then the roots will be real and equal



    0 = 5.887 \times 10^{24}x^2 - 4.648 \times 10^{33}x + 9.059 \times 10^{41} = 0

    Divide by 10^{24}

    5.887x^2 - 4.648 \times 10^{9}x + 9.059 \times 10^{17} = 0
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    Not always. If b^2-4ac=0 then the roots will be real and equal
    Hm...okay. Well, could some one try this and tell me what you get. I got "x" to be 4.648E33 for both the negative and the positive answer. This is also one of my numbers up there...something seems wrong. This is why I came here, because I am almost sure I am doing something wrong.
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  8. #8
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    Quote Originally Posted by zach610 View Post
    Hm...okay. Well, could some one try this and tell me what you get. I got "x" to be 4.648E33 for both the negative and the positive answer. This is also one of my numbers up there...something seems wrong. This is why I came here, because I am almost sure I am doing something wrong.
    I am guessing because b >> b^2-4ac then you can assume b^2-4ac=0 which would give a root equal to -\frac{b}{2a}
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  9. #9
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    Quote Originally Posted by e^(i*pi) View Post
    I am guessing because b >> b^2-4ac then you can assume b^2-4ac=0 which would give a root equal to -\frac{b}{2a}
    Well, when I try that, I get a completely different number. This is exactly why I posted this. This problem is much harder than I thought. Not sure where to go from here.
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