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Math Help - Simplify

  1. #1
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    Simplify

    I have the equation x^2-x=-147.
    I played around with it but couldn't find away to get a definite answer without having to use trial and error.
    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    I have the equation x^2-x=-147.
    I played around with it but couldn't find away to get a definite answer without having to use trial and error.
    Thanks!
    there is good reason you had problems. this has no real solutions

    if you bring everything to one side: x^2 - x + 147 = 0

    and apply the quadratic formula: x = \frac {1 \pm \sqrt{1 - 4(147)}}2

    obviously what is under the square root is negative, thus we have complex solutions.
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    there is good reason you had problems. this has no real solutions

    if you bring everything to one side: x^2 - x + 147 = 0

    and apply the quadratic formula: x = \frac {1 \pm \sqrt{1 - 4(147)}}2

    obviously what is under the square root is negative, thus we have complex solutions.

    But wouldn't x^2 - x + 147 = 0 be in standard form so that it could be factored? (x-?)(x+?)
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Morzilla View Post
    But wouldn't x^2 - x + 147 = 0 be in standard form so that it could be factored? (x-?)(x+?)
    Technically yes.

    x^2 - x + 147 = \left ( x - \frac{1}{2} - i~\frac{\sqrt{587}}{2} \right ) \left ( x - \frac{1}{2} + i~\frac{\sqrt{587}}{2} \right )

    As Jhevon pointed out, there are no real solutions to this problem. And the factorization is awful looking. Stick with the quadratic formula for this one; not every quadratic factors nicely.

    -Dan
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  5. #5
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    ohhhhh, so now my question is, how did we go from <br />
x^2 - x + 147 = 0 this to = \frac {1 \pm \sqrt{1 - 4(147)}}2

    what are the steps basically?

    I must admit, this is getting interesting, and it's not even my problem!?!

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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Morzilla View Post
    ohhhhh, so now my question is, how did we go from <br />
x^2 - x + 147 = 0 this to = \frac {1 \pm \sqrt{1 - 4(147)}}2

    what are the steps basically?

    I must admit, this is getting interesting, and it's not even my problem!?!

    I used the quadratic formula. it is derived by completing the square. basically it says, for any quadratic of the form y = ax^2 + bx + c, the roots of the quadratic (that is, the solutions to ax^2 + bx + c = 0) are given by:

    x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    y = ax^2 + bx + c, the roots of the quadratic (that is, the solutions to ax^2 + bx + c = 0) are given by:

    x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}
    ok I remember and understand y = ax^2 + bx + c and making all equal to zero, or as you mentioned, but I still can't understand the last part...
    x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}

    I may never be able to get it
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  8. #8
    Senior Member DivideBy0's Avatar
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    It's derivation is quite nice:

    Quadratic Formula Derivation

    But you will need to know how to complete the square. It is a necessary prerequisite.

    This is hard stuff, so take your time with it and don't get yourself down if you can't understand it after a while.
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  9. #9
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    Here I attach another solution.
    Attached Files Attached Files
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  10. #10
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    Quote Originally Posted by Krizalid View Post
    Here I attach another solution.
    Cool! gracias pero, where did 4a and the b^2 come from!?
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