- #1

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^{2}- 8ab + 3b

^{2}=0?

I know there are general formulae, but I'm not sure how to use them:

a

^{2}-2ab+b^

^{2}= (a-b)

^{2}

a

^{2}+2ab+b^

^{2}= (a+b)

^{2}

a

^{2}-b

^{2}= (a+b)(a-b)

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- B
- Thread starter vetgirl1990
- Start date

- #1

- 85

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I know there are general formulae, but I'm not sure how to use them:

a

a

a

- #2

berkeman

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Is this for schoolwork?^{2}- 8ab + 3b^{2}=0?

I know there are general formulae, but I'm not sure how to use them:

a^{2}-2ab+b^^{2}= (a-b)^{2}

a^{2}+2ab+b^^{2}= (a+b)^{2}

a^{2}-b^{2}= (a+b)(a-b)

- #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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- #5

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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.Is this for schoolwork?

Perhaps I should have posted it in the homework questions folder?

- #6

member 587159

Let us know when you find the answer :)

- #7

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I'm not quite sure I understand what you mean.Consider it as a second grade equation in a in the form: da^2 + ea + f = 0. Solve for a.

- #8

berkeman

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Yeah, I didn't understand his reply either.I'm not quite sure I understand what you mean.

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. Definitely all homework-like questions should be posted in the Homework Help forums, though.Perhaps I should have posted it in the homework questions folder?

- #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

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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:

- #11

- 138

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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!!)

- #12

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Please see my addition to post #10.

- #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

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I miscopied the -6ab". It is a "-8ab". See my correction (edited). 4a^2-8ab+3b^2 factors quite easily.Indeed, no answer in ℝ

- #15

berkeman

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See the LaTeX turorial in the Help/How-To list under INFO at the top of the page.(note to self.. must learn how to post powers!!)

- #16

berkeman

Mentor

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How do you solve 4a2 - 8ab + 3b2=0?

It is indeed. Now let's let the OP try to figure it out please...It is a polynomial that happens to factor.

- #17

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3x^2 -8rx +4r^2

=3x^2 -2rx -6rx +4r^2

=x(3x-2r) -2r(3x-2r)

=(3x-2r)(x-2r)

:)

=3x^2 -2rx -6rx +4r^2

=x(3x-2r) -2r(3x-2r)

=(3x-2r)(x-2r)

:)

- #18

SammyS

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Do you possibly mean second degree?Consider it as a second grade equation in a in the form: da^2 + ea + f = 0. Solve for a.

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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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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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.)

Q1 Did anyone solve it?

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

Thanks

Matt

But the opening line in the OP asks about solving 4a

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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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.

- #23

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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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One of the reasons you will seldom see me in the maths forum, is because I don't speak maths.

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

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.

- #25

SammyS

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

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

Alternatively, divide your equation by b

##\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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