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Math Help - Solving quadratic equations

  1. #1
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    Solving quadratic equations

    If I solve a quadratic equation and I get the values for x as
    x=1.788887778 and x=1.11382983
    what would they usually round to?


    is this allowed?
    Thanks
    Last edited by wolfhound; November 9th 2009 at 08:20 AM.
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  2. #2
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    Hi,

    The values of x that you find are only approximative. The goal of solve a quadratic equation is to find two, one (equals) or zero (in \mathbb{R}, of course).

    Don't try to put they back to the equation, it doesn't work if the values aren't integers.
    Last edited by mr fantastic; November 9th 2009 at 07:41 PM. Reason: Fixed latex
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  3. #3
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    Hi but I need to put them back in to check if they are the correct answer,

    could you tell me if I have the correct answers

    I had to solve using square root formula
    Last edited by wolfhound; November 9th 2009 at 08:20 AM.
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  4. #4
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    Talking

    Quote Originally Posted by wolfhound View Post
    If I solve a quadratic equation and I get the values for x as x=1.78 and x=1.11
    These would be the solutions only if the original quadratic were x^2 - 2.89x + 1.9758, or some multiple thereof. Was it?

    Quote Originally Posted by wolfhound View Post
    ...so they round to x=1.8 and x=1.1
    On what basis had you concluded that the solutions were to be rounded to one decimal place?

    Quote Originally Posted by wolfhound View Post
    ...and I input the values back in
    3(-1.1)^2-2(-1.1)-6=0
    Since you are not plugging the actual solutions back into the original equation, you cannot check to see if they are correct. At best, you can confirm that they are at least close.

    FYI: Non-integral values can be checked. But you'd have to work exactly. For instance, 3x^2 - 2x - 6 = 0 solves as:

    x\, =\, \frac{1\, \pm\, \sqrt{19}}{3}

    To check one of the solutions:

    3\left(\frac{1\, +\, \sqrt{19}}{3}\right)^2\, -\, 2\left(\frac{1\, +\, \sqrt{19}}{3}\right)\, -\, 6

    3\left(\frac{20\, +\, 2\sqrt{19}}{3}\right)\, -\, 2\left(\frac{1\, +\, \sqrt{19}}{3}\right)\, -\, 6

    \frac{20\, +\, 2\sqrt{19}}{3}\, -\, \frac{2\, +\, 2\sqrt{19}}{3}\, -\, 6

    \frac{20}{3}\, +\, \frac{2\sqrt{19}}{3}\, -\, \frac{2}{3}\, -\, \frac{2\sqrt{19}}{3}\, -\, 6

    \frac{18}{3}\, -\, 6

    ...and so forth.
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  5. #5
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    Hi thanks but
    I am only allowed use the -b (square root )b^2-4ac/2a formula or standard factoring!
    I got these values from -b formula
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  6. #6
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    would my answer be correct to the - b formula???
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  7. #7
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    Quote Originally Posted by wolfhound View Post
    Hi thanks but
    I am only allowed use the -b (square root )b^2-4ac/2a formula or standard factoring!
    I got these values from -b formula
    I don't see what that has to do with anything anyone has said here. And you have not answered a very important question- why did you round your answers to one decimal place?

    Apparently, your original equation was 3x^2- 2x- 6= 0, since you put x= -1.1 (after saying that x= 1.11 was a solution!) into that.

    By the quadratic formula, x= \frac{2\pm\sqrt{4+ 72}}{6}= \frac{2\pm\sqrt{76}}{6}. x= \frac{2+ \sqrt{76}}{2} and x= \frac{2- \sqrt{76}}{2} are the correct roots.

    \sqrt{76} is about 8.72 so x is approximately 1.79 or -1.14. I've reduced those to two decimal places because you gave two decimal places as your first answer. What reason do you have to further reduce to 1.8 and -1.1?
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  8. #8
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    Hi,
    I thought I was supposed to round to 2 digits because 76(two digits) was the largest number
    Could you tell me what the correct rounding would be PLEASE
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  9. #9
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    round

    the exact rounded values are 1.79 and -1.12.
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  10. #10
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    Quote Originally Posted by wolfhound View Post
    Hi,
    I thought I was supposed to round to 2 digits because 76(two digits) was the largest number
    Could you tell me what the correct rounding would be PLEASE
    The correct rounding to use is the rounding that the question tells you to do!! Otherwise, give exact surd values. Until you state the entire question, exactly as it was written and with all instructions, there is nothing more can be said.
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  11. #11
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    Quote Originally Posted by wolfhound View Post
    Hi,
    I thought I was supposed to round to 2 digits because 76(two digits) was the largest number
    Could you tell me what the correct rounding would be PLEASE
    Well, you are wrong. In working with measurements, your accuracy cannot be greater than that of the least accurate figure- the one with the fewest significant digits, not most. But there are no measurements involved here so that is irrelevant.
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