Finding Solutions for a Quadratic Equation with Two Variables

[vetgirl1990]

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)

 

[berkeman]

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?

 

[member 587159]

Consider it as a second grade equation in a in the form: da^2 + ea + f = 0. Solve for a.

 

[Charles Link]

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.

 

[vetgirl1990]

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?

 

[member 587159]

Let us know when you find the answer :)

 

[vetgirl1990]

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.

 

[berkeman]

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. Definitely all homework-like questions should be posted in the Homework Help forums, though.

 

[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?

 

[Charles Link]

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.

 

[mgkii]

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!)

 

[Charles Link]

Please see my addition to post #10.

 

[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

 

[Charles Link]

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.

 

[berkeman]

(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.

 

[berkeman]

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...

 

[vetgirl1990]

3x^2 -8rx +4r^2
=3x^2 -2rx -6rx +4r^2
=x(3x-2r) -2r(3x-2r)
=(3x-2r)(x-2r)

:)

 

[SammyS]

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 .

 

[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.

 

[mgkii]

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

 

[SammyS]

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.

 

[OmCheeto]

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.

 

[mgkii]

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

 

[OmCheeto]

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.

 

[SammyS]

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.)

 

[Charles Link]

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

The OP essentially factored the 4a^2-8ab+3b^2 in post #17, replacing "a" with "r" and "b" with "x". (The OP's answer also has each of the two terms multiplied by "-1" ; it's not necessary for the "a" terms to get the "+" sign. It's equally correct for the "b's " to get the "+" sign.)

 

[AlexS]

You know that you can get any factorization you want with any formula? Yes, you can use general formula such as (a+b)^2=a^2+b^2+2ab, but you can also find any factorization you want for any formula, using Horner's method for factorization (https://en.wikipedia.org/wiki/Horner's_method). Now, if you have problems with the concept, tell me.

 

[AlexS]

Now, I come back with another answer because I understand that you need to find the solutions of this equation. It is actually impossible because you can't. It is an equation with 2 variables, and to understand why you can't , for instance, when we have a system with 2 variables, we need 2 EQUATIONS to find the solution of this system, with real solutions. If you want to find integer solutions, that is another problem. If you want to find integer solutions, you obly have to do a factorization and you get the answer. I think that in your problem should be more about a or b. The only thing you can get with this equations is that b=2a or b=2a/3. That's all I can see.