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Math Help - Real VAlues

  1. #1
    Member Rimas's Avatar
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    Real VAlues

    Find all real values of x that satisfy (x^2-5x+5)^x^2-9x+2=1
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Rimas View Post
    Find all real values of x that satisfy (x^2-5x+5)^x^2-9x+2=1
    Okay, we know that (or i hope we know that) anything raised to the zero gives us one. So that means our power x^2 - 9x + 2 = 0. Now we just solve the quadratic equation.

    x^2 - 9x + 2 = 0
    Using the quadratic formula:
    => x = [9 +/- sqrt{9^2 - 4(1)(2)}]/2
    = [9 +/- sqrt(73)]/2
    = 8.772 or 0.228

    Plugging these into your calculator won't give you exactly zero, since they are decimal approximations, but you get -0.000016 or something like that, which is good enough. If you want an answer closer to zero, use more decimal points. We also want to double check these values for x into the base, that is into x^2 - 5x + 5, to see if they cause that to be zero. Since 0^0 is meaningless (i think), we would disregard any solution that is also a root of the base, both these solutions are ok.
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