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Math Help - Solving by completing the square.

  1. #1
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    Talking Solving by completing the square.

    Hey I need help on figuring out this problem...and I'm in middle school doing algebra made for people in high school!!! Ok here is the problem:
    x-4x=17
    I know to you that probably looks easy, but what do you do????
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  2. #2
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    to complete the square

    x^2+ax to complete the square we take half the coeffient on the x term.

    So in this case \frac{a}{2} and then we square it

    \left( \frac{a}{2} \right)^2=\frac{a^2}{4}

    Then we add this to get..

    x^2+ax+\frac{a^2}{4}=(x+\frac{a}{2})^2

    Now to apply this to your problem.

    x^2-4x=17 so we take half the x term and square it \left( \frac{-4}{2} \right)^2=4

    So now we add 4 to both sides of the equation

    x^2-4x +4=17+4 \iff (x-2)^2=21 now take the square root of both sides

    x-2=\pm \sqrt{21} \iff x=2 \pm \sqrt{21}

    Good luck
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  3. #3
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    x-4x=17

    Now , to factorize by completing square, we must make the LHS in the form
    ( a - b )^2 that is a^2 -2ab + b^2

    It is found that if we add 4 to the equation then the LHS does become of the above form

    so adding 4 we have,

    x^2 - 4x + 4 = 17 + 4

    therefore

    (x)^2 - 2(x)(2) + (2)^2 = 21

    so, factorizing we get,

    (x-2)^2 = 21, transposing 21 we get

    [(x-2)]^2 - 21 = 0

    This looks similar, infact it imitates the identity , a^2 - b^2 = (a-b)(a+b)

    so further factorizing it we get,

    [x-2+ ROOT(21) ] [x-2- ROOT(21)]=0

    SOLUTIONS, either x = 2- ROOT(21) OR x = 2 + ROOT(21)

    hence found.
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