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Math Help - [SOLVED] Hard Algebra Questions

  1. #1
    ksle82
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    [SOLVED] Hard Algebra Questions

    Simplify the followings:

    A) (a^4 + b^4)/(a^2 +b^2)

    B) (2 + 2i)/(3-2i), where i is the imaginary
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ksle82
    Simplify the followings:

    A) (a^4 + b^4)/(a^2 +b^2)

    B) (2 + 2i)/(3-2i), where i is the imaginary
    For the way most would want the answer, A) is already as simple as it is going to get.

    In B) we want to remove the radical in the denominator. So multiply top and bottom by the complex conjugate of the denominator:
    \frac{2+2i}{3-2i} \cdot \frac{3+2i}{3+2i}

    Since i^2=-1:
    =\frac{6+4i+6i+4i^2}{9+6i-6i-4i^2}=\frac{6+10i-4}{9+4}

    =\frac{2+10i}{13}

    -Dan
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  3. #3
    MHF Contributor
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    Quote Originally Posted by ksle82
    Simplify the followings:

    A) (a^4 + b^4)/(a^2 +b^2)

    B) (2 + 2i)/(3-2i), where i is the imaginary
    Here are some ways:

    A) (a^4 +b^4) / (a^2 +b^2) --------(1)

    A.1) By long division, = (a^2 -b^2) +(2b^4)/(a^2 +b^2) -----------------------answer.

    A.2) Get a factor (a^2 +b^2) from the numerator (a^4 +b^4):
    Squaring the denominator,
    (a^2 +b^2)^2 = a^4 +2(a^2)(b^2) +b^4
    (a^2 +b^2)^2 = (a^4 +b^4) +2(a^2)(b^2)
    (a^4 +b^4) = (a^2 +b^2)^2 -2(a^2)(b^2) ------(i)

    Substitute that into (1),
    (a^4 +b^4) / (a^2 +b^2) --------(1)
    = [(a^2 +b^2)^2 -2(a^2)(b^2)] / (a^2 +b^2)
    = (a^2 +b^2) -2(a^2)(b^2)/(a^2 +b^2) --------------answer.

    =============================
    B) (2 +2i) / (3 -2i) -----------------------(2)

    Make the denominator a real number, muliply both numerator and denominator by the conjugate of the denominator, that is by (3 +2i):

    = [(2 +2i)(3 +2i)] / [(3 -2i)(3 +2i)]
    = [6 +4i +6i +4i^2] / [9 -4i^2]
    Since i^2 = -1, then,
    = [2 +10i] / [13]
    = (1/13)(2 +10i)
    = (2/13)(1 +5i) -----------answer.
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