Factor an equation second order in ##x,y## with other variables

  • Thread starter Thread starter brotherbobby
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
The discussion revolves around factoring a second-order equation involving variables x, y, and constants a, b, and c. Participants express their struggles with the problem and share various approaches, including expanding expressions and completing the square. There is a consensus on the challenge of solving the equation without relying on computational tools like Wolfram Alpha, as they prefer traditional methods for learning. Suggestions include exploring the relationships between coefficients and using known algebraic identities to find factors. The conversation highlights a blend of frustration and collaborative problem-solving in tackling complex algebraic expressions.
brotherbobby
Messages
749
Reaction score
169
Homework Statement
Resolve into two or more factors : ##\large{\boldsymbol{2a^2x^2-2(3b-4c)(b-c)y^2+abxy}}##.
Relevant Equations
1. ##a^2-b^2=(a+b)(a-b)##
1706337399659.png
Statement of the problem :
Let me copy and paste to the right the problem as it appears in the text.

Attempt : I couldn't go far into the solution. Below is my hopeless attempt.

1706337172892.png


Request : Any hints would be welcome.

[There are no solutions provided, but the answers are at the back of the text. I don't want to cheat my way out by looking at the factorised answer and then working my way back, though I admit it can lead to some learning too in the process]
 
Physics news on Phys.org
Hmm not sure, wolfram does this very fast and it gives 2 factors with 3 terms each that is $$(A+B+C)(D+E+F)$$
 
Thanks but how to go about it? I have the answer already at the back of the text, am sure, but that doesn't help. :-(
 
Well I don't know, looking at the wolfram answer the terms are +-(ax) +-(by) +-(cy), so I guess something like full expand the expression and then try to identify these terms and/or their squares.
 
The website is good as a check, but not one to learn mathematics from.
I'd like to do it the old way, not least because my text is from the 1800's!
 
  • Like
  • Informative
Likes DaveE, fresh_42 and Delta2
My original idea before looking at wolfram was to write $$(3b-4c)(b-c)y^2=3(b-c)^2y^2-c(b-c)y^2$$ and see if that helps but I abandon it maybe this idea inspires you but I very much doubt it.
 
Why so? Not bad at all. Let me give it a try.
I am clueless with this problem so every little helps.
 
brotherbobby said:
Homework Statement: Resolve into two or more factors : ##\large{\boldsymbol{2a^2x^2-2(3b-4c)(b-c)y^2+abxy}}##.
Relevant Equations: 1. ##a^2-b^2=(a+b)(a-b)##

View attachment 339261Statement of the problem : Let me copy and paste to the right the problem as it appears in the text.

Attempt : I couldn't go far into the solution. Below is my hopeless attempt.

View attachment 339259

Request : Any hints would be welcome.

[There are no solutions provided, but the answers are at the back of the text. I don't want to cheat my way out by looking at the factorised answer and then working my way back, though I admit it can lead to some learning too in the process]
Why not multiply back your factored form see if you get back your initial expression?
 
What comes to mind is "completing the square". We have ...
\begin{align*}
2a^2x^2-2(3b-4c)(b-c)y^2+abxy&=2a^2x^2-(6b^2-14bc+8c^2)y^2+abxy\\
&=2a^2x^2-(6b^2-14bc+(49/6)c^2)y^2+abxy + (1/6)c^2y^2\\
&=2a^2x^2 -(\sqrt{6}b-(7c/\sqrt{6}))^2 +abxy + (1/6)c^2y^2\\
&\ldots
\end{align*}
... but I admit that I don't have a clue how to go on.

Another idea is to use only the terms with a coefficient ##a## first ...
\begin{align*}
2a^2x^2-2(3b-4c)(b-c)y^2+abxy&=\left(\sqrt{2}ax+\dfrac{by}{2\sqrt{2}}\right)^2-\dfrac{b^2}{8}y^2 \ldots\\
\phantom{2a^2x^2-2(3b-4c)(b-c)y^2+abxy=}&\ldots -\left(\sqrt{6}b-\dfrac{7c}{\sqrt{6}}\right)^2 + \dfrac{c^2}{6}y^2\\
&=\ldots
\end{align*}
... which looks as if ##(A-B)(A+B)=A^2-B^2## could now be applied twice.

This exercise looks like it is meant to practice all variations of the formulas ##(A\pm B)(A\mp B)=\ldots ## and completing the square.
 
Last edited:
  • Like
Likes MatinSAR and Delta2
  • #10
Cant see the full solution in wolfram cause I don't have subscription there but it claims as first step to split the product $$4a^2y^2(3b-4c)(b-c) $$into two factors that have sum ##aby##. Those AI engines can be amazing but I don't think a human could think that unless of course he was thinking this for like days.
 
  • #11
Delta2 said:
Cant see the full solution in wolfram cause I don't have subscription there but it claims as first step to split the product $$4a^2y^2(3b-4c)(b-c) $$into two factors that have sum ##aby##. Those AI engines can be amazing but I don't think a human could think that unless of course he was thinking this for like days.
That human didn't need to think. I guess he just expand sth like (ax+b+c)(ax+y-c) to get sth else and asks us to find first two factors. I can't see how it helps to learn sth it's more similar to a puzzle.

I don't know why Latex doesn't work in my posts anymore ...
 
  • #12
MatinSAR said:
I don't know why Latex doesn't work in my posts anymore ...
It requires refreshing the page occasionally.
 
  • #13
MatinSAR said:
That human didn't need to think. I guess he just expand sth like (ax+b+c)(ax+y-c) to get sth else and asks us to find first two factors. I can't see how it helps to learn sth it's more similar to a puzzle.

I don't know why Latex doesn't work in my posts anymore ...
as human here you mean the problem setter? I was talking about the guy that tries to solve this problem.
 
  • #14
fresh_42 said:
It requires refreshing the page occasionally.
Thanks.
Delta2 said:
as human here you mean the problem setter?
Yes. I misunderstood your idea. Sorry.
 
  • #15
Given the coefficient of x^2 is 2a^2, it makes sense to look for a solution <br /> 2(ax + By)(ax + Dy) = 2a^2x^2 + 2a(B + D)xy + 2BDy^2. Since p = B + D and q = BD are known, we can find B and D as the roots of <br /> z^2 - pz + q = 0.

Since those factors involve square roots, we can instead try (2ax + By)(ax + Dy) = 2a^2x^2 + a(B + 2D)xy + BDy^2. which leads to B + 2D = b and BD = -2(3b - 4c)(b - c) which by inspection has a solution B = -(3b - 4c), D = 2(b - c).
 
Last edited:
  • Like
  • Informative
Likes DaveE, e_jane, MatinSAR and 1 other person
Back
Top