1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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,

    4[(1/4)y^2-(3/4)xy+(9/16)y^2]-4(9/16)y^2+10y^2

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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

    epenguin

    User Avatar
    Homework Helper
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook