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B How to solve: a^2 - ab + b^2

  1. Apr 22, 2016 #1
    How do you solve 4a2 - 8ab + 3b2=0?

    I know there are general formulae, but I'm not sure how to use them:
    a2-2ab+b^2 = (a-b)2
    a2+2ab+b^2= (a+b)2
    a2-b2 = (a+b)(a-b)
     
  2. jcsd
  3. Apr 22, 2016 #2

    berkeman

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    Is this for schoolwork?
     
  4. Apr 22, 2016 #3

    Math_QED

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    Consider it as a second grade equation in a in the form: da^2 + ea + f = 0. Solve for a.
     
  5. Apr 22, 2016 #4

    Charles Link

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    I worked something similar a few years back=perhaps you will find this relevant. a^2 +2ab +b^2 is a perfect square. It can always be written as (a+b)^2. The question I had (probably since algebra class), but never explored, was could a^2+ab+b^2 ever be a perfect square? i.e. a^2+ab+b^2=c^2 where a, b, and c are integers. I did find some solutions. I expect you could do the same for a^2-ab+b^2. In general, there are only a couple expressions like a^2+2ab+b^2 and a^2-b^2 that factor in such a simple manner. You can even do the same thing for a^2+b^2=c^2 and there are a number of solutions like the 3,4, 5 and 5,12, 13 right triangles. (You should recognize the Pythagoren theorem equation a^2+b^2=c^2 for this latter case.) And of course, any integral multiples of these solutions will also be solutions.
     
  6. Apr 22, 2016 #5
    Partially. I'm just studying for a physics exam. While I understand the physics behind this particular question, I'm just stuck on how to solve for the final answer from the above expression.

    Perhaps I should have posted it in the homework questions folder?
     
  7. Apr 22, 2016 #6

    Math_QED

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    Let us know when you find the answer :)
     
  8. Apr 22, 2016 #7
    I'm not quite sure I understand what you mean.
     
  9. Apr 22, 2016 #8

    berkeman

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    Yeah, I didn't understand his reply either.

    You need to factor the equation to find the roots. Have you learned much about how to factor polynomials yet?

    It sounds more like a general question, so it's probably okay here for now. :smile: Definitely all homework-like questions should be posted in the Homework Help forums, though.
     
  10. Apr 22, 2016 #9

    Math_QED

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    Do you know how to solve the equation: ax^2 + bx+ c = 0 for x?
     
  11. Apr 22, 2016 #10

    Charles Link

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    One additional item: The form a^2-ab+b^2 appears in the factoring of a^3+b^3=(a+b)(a^2-ab+b^2). I think many people ask themselves upon seeing the expression a^2-ab+b^2 here, does it factor further, and the answer is no. And your other question, how do you factor (4a^2-8ab +3b^2)? It is a polynomial that happens to factor. That one you should be able to get an answer for.
     
    Last edited: Apr 22, 2016
  12. Apr 22, 2016 #11
    Ok I'll bite, but not sure I get any closer :-)

    Sticking it through the old x = -b +/- root(b squared..... etc

    a = b + or - 2b squared

    (note to self.. must learn how to post powers!!)
     
  13. Apr 22, 2016 #12

    Charles Link

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    Please see my addition to post #10.
     
  14. Apr 22, 2016 #13

    Math_QED

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    Indeed, no answer in ℝ
     
  15. Apr 22, 2016 #14

    Charles Link

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    I miscopied the -6ab". It is a "-8ab". See my correction (edited). 4a^2-8ab+3b^2 factors quite easily.
     
  16. Apr 22, 2016 #15

    berkeman

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    See the LaTeX turorial in the Help/How-To list under INFO at the top of the page. :smile:
     
  17. Apr 22, 2016 #16

    berkeman

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    It is indeed. Now let's let the OP try to figure it out please... :smile:
     
  18. Apr 22, 2016 #17
    3x^2 -8rx +4r^2
    =3x^2 -2rx -6rx +4r^2
    =x(3x-2r) -2r(3x-2r)
    =(3x-2r)(x-2r)

    :)
     
  19. Apr 22, 2016 #18

    SammyS

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    Do you possibly mean second degree?

    I don't know any second graders who can do this. I'm sure that they are rare if not non-existent .
     
  20. Apr 23, 2016 #19

    Math_QED

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    Oh yes second degree. I'm sorry English is not my native language.
     
  21. Apr 25, 2016 #20
    So how does this factor? I got as far as I could with the traditional quadratic formula but I'm pretty sure it was a fail!

    Q1 Did anyone solve it?
    Q2 If so... Was it by intuition/brute force; or is there a reusable technique?

    Thanks
    Matt
     
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