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Math Help - Looking for a Quadratic tip(s)

  1. #1
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    Looking for a Quadratic tip(s)

    I found a quadratic calculator online, so to an extent this is a non-question. Also I have been doing this problem as a sort of hobby, so I'm much more interested in the process then the solution.

    I have been trying to describe the intersection of two lines. After several pages of calculations I have ended up with a quadratic something like;
    .396260075y^2+1201.957114y+800.3476=0

    First off I assume that problems like this needed too be completed back in ye good ole days. And I am wondering if there is anything like the Rule of Sarrus, which is just a trick to make solving a matrix easier when doing it by hand.

    Secondly I was wondering what kind of research has been done on rounding versus not rounding. Like say in the above quadratic, I just droped most of the decimal positions too make the calculations easier, what would this do to the two answers? (It seems like this sort of thing would be important in Engineering but I have no idea.)

    Also, just to make sure I'm understanding this quadratic calculator correctly. It's output is a number that makes each binomial term equal to 0, after factoring?
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  2. #2
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    Quote Originally Posted by bkbowser View Post
    Also, just to make sure I'm understanding this quadratic calculator correctly. It's output is a number that makes each binomial term equal to 0, after factoring?
    Should be 2 outputs.

    Lots of those calculators:
    quadratic calculator online - Google Search
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  3. #3
    PQR
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    Quote Originally Posted by bkbowser View Post
    It's output is a number that makes each binomial term equal to 0, after factoring?
    Two outputs each of which makes one binomial term goes to zero.
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  4. #4
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    Quote Originally Posted by bkbowser View Post
    .396260075y^2+1201.957114y+800.3476=0
    No (real) difference between that and (as example) 2y^2 - 7x + 3 = 0;
    online calculator will output TWO numbers: 1/2 and 3,
    hence (2x - 1)(x - 3) = 0.

    As far as getting an approximate, seems to be up to whoever
    wants the answer; you could multiply by 10, then round up:
    4y^2 + 12020y + 8004 = 0 ; then divide that by 4:
    y^2 + 3005y + 2001 = 0
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