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    i worked this question out before but now for some reason i dont know how to do it now
    the question is solve 4x^2-8=0 using differnce of two squares

    there's also another one, solve 2x^2-8x-11=0 using the completed square method
    HELP!
    thank you :o:
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    2x^2 - 8x - 11 = 0

    deal with the first two terms only

    2x^2 - 8x = 0

    factorise out the 2

    2(x^2 - 4x) = 0

    square the bracket

    2((x - 2)^2) = 0

    subtract the number in the bracket and square it

    2((x - 2)^2 - 2^2) = 0

    bring down the constant

    (x - 2)^2 - 8 - 11 = 0

    (x - 2)^2 - 19 = 0
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    (Original post by LazyWorseThanInfidel)
    2x^2 - 8x - 11 = 0

    deal with the first two terms only

    2x^2 - 8x = 0

    factorise out the 2

    2(x^2 - 4x) = 0

    square the bracket

    2((x - 2)^2) = 0

    subtract the number in the bracket and square it

    2((x - 2)^2 - 2^2) = 0

    bring down the constant

    (x - 2)^2 - 8 - 11 = 0

    (x - 2)^2 - 19 = 0
    All of these bolded lines are wrong. I can see what you're doing, and it's the right method, but please don't mislead people into thinking that these things are actually equal to 0. What you should be doing is:

    2x^2 - 8x - 11 = 0
    (2x^2 - 8x) - 11 = 0
    2(x^2 - 4x) - 11 = 0
    2[(x - 2)^2 - 2^2] - 11 = 0

    ...and for some reason, in the two underlined lines, you dropped the factor of 2.
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    sorry il have to step it up a league well spotted
 
 
 
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