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Factoring x and y in quadradic form

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data
    factor 4x^2-6xy+10y^2

    2. Relevant equations

    3. The attempt at a solution

    completing the square i get,


    but i don't know what to do with the bracketed term. It cant be factored easily
  2. jcsd
  3. Feb 24, 2010 #2


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    Science Advisor

    I presume that is supposed to be [itex]x^2[/itex], not [itex]y^2[/itex] at the first but then I have no idea why you have "1/4"

    [itex]4x^2- 6xy+ 10y^2= 4(x^2- (3/2)xy)+ 10y^2[/itex]
    [itex]= 4(x^2- (3/2)xy+ (9/16)y^2- (9/16)y^2)+ 10y^2[/itex]
    [itex]= 4(x^2- (3/2)xy+ (9/16)y^2)- (9/4)y^2+ 10y^2[/itex]
    [itex]= 4(x- (3/4)y)^2+ (31/4)y^2[/itex]
    In order to factor that we need a difference of two squares so write it as
    [itex]= 4(x- (3/4)y)^2- (-31/4)y^2[/itex]

    [itex]= 4(x- (3/4)y)^2- ((\sqrt{31}i/2)y)^2[/itex]

    [tex]= (2(x- (3/4)y)- (i\sqrt{31}/2)y)(2(x-(3/4)y)+ (i\sqrt{31}/2)y)[/tex]
  4. Feb 24, 2010 #3


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    Gold Member

    Are you sure you transcribed it right?:devil:

    If it were 4x^2-6xy-10y^2 it would all work out nice.

    The people who invent these probs tend to avoid such :yuck: things as this is giving - it is easier for them too. :wink:

    But then maybe they also sometimes set traps.

    Easy to check.
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