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

  • #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)
 

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  • #2
berkeman
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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)
Is this for schoolwork?
 
  • #3
member 587159
Consider it as a second grade equation in a in the form: da^2 + ea + f = 0. Solve for a.
 
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  • #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.
 
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  • #5
Is this for schoolwork?
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?
 
  • #6
member 587159
Let us know when you find the answer :)
 
  • #7
Consider it as a second grade equation in a in the form: da^2 + ea + f = 0. Solve for a.
I'm not quite sure I understand what you mean.
 
  • #8
berkeman
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I'm not quite sure I understand what you mean.
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?

Perhaps I should have posted it in the homework questions folder?
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.
 
  • #9
member 587159
I'm not quite sure I understand what you mean.

Do you know how to solve the equation: ax^2 + bx+ c = 0 for x?
 
  • #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.
 
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  • #11
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I'm not quite sure I understand what you mean.

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
member 587159
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^2-6ab +3b^2)? It is a polynomial that happens to factor. That one you should be able to get an answer for.

Indeed, no answer in ℝ
 
  • #14
Charles Link
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Indeed, no answer in ℝ
I miscopied the -6ab". It is a "-8ab". See my correction (edited). 4a^2-8ab+3b^2 factors quite easily.
 
  • #15
berkeman
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(note to self.. must learn how to post powers!!)
See the LaTeX turorial in the Help/How-To list under INFO at the top of the page. :smile:
 
  • #16
berkeman
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How do you solve 4a2 - 8ab + 3b2=0?
It is a polynomial that happens to factor.
It is indeed. Now let's let the OP try to figure it out please... :smile:
 
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  • #17
3x^2 -8rx +4r^2
=3x^2 -2rx -6rx +4r^2
=x(3x-2r) -2r(3x-2r)
=(3x-2r)(x-2r)

:)
 
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  • #18
SammyS
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Consider it as a second grade equation in a in the form: da^2 + ea + f = 0. Solve for a.
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 .
 
  • #19
member 587159
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 .

Oh yes second degree. I'm sorry English is not my native language.
 
  • #20
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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
 
  • #21
SammyS
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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
Well, there's the thread title. Seems to asks about factoring a^2 - ab + b^2, or solving a^2 - ab + b^2 = 0. (It doesn't fully state either.)

But the opening line in the OP asks about solving 4a2 - 8ab + 3b2 = 0 .

The left hand side of the latter equation can be factored to get the solution.

The expression in the title cannot be factored.
 
  • #22
OmCheeto
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I cheated and used Wolfram Alpha to get the answers.
But that was only so I could make sure I did my maths right.
It's been decades since I've studied maths, so I wasn't sure if solving the problem with the quadratic method would work.
To my surprise, it did work. :partytime:
 
  • #23
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I'm still missing something here, please bear with me ....

I also used the quadratic formula, but it gives solutions for a that includes factors of b. How did you move to simple factors?

Thanks
 
  • #24
OmCheeto
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I'm still missing something here, please bear with me ....

I also used the quadratic formula, but it gives solutions for a that includes factors of b. How did you move to simple factors?

Thanks
One of the reasons you will seldom see me in the maths forum, is because I don't speak maths.
I don't know what "move to simple factors" means.
Sorry.

ps. My answers, a=f(b), were not the same as Wolfram Alpha's, b=f(a). But it only took me about 3 seconds to reconcile. :smile:
 
  • #25
SammyS
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I'm still missing something here, please bear with me ....

I also used the quadratic formula, but it gives solutions for a that includes factors of b. How did you move to simple factors?

Thanks
If you treat a as the variable, then yes, it's solution will include b .

Alternatively, divide your equation by b2 giving:
##\displaystyle 4\left(\frac ab \right)^2-8\left(\frac ab \right)+3=0 ##​

Solve that for ##\displaystyle \ \frac ab \ ## either by factoring, or by the quadratic formula. (Yes, this is factorable.)
 

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