How do I complete the square for x^2 + y^2 - 2pxy?

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Discussion Overview

The discussion revolves around the process of completing the square for the expression x^2 + y^2 - 2pxy. Participants seek clarification on the steps involved in this algebraic technique, particularly in the context of a classroom assignment.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • Some participants propose that the expression can be rewritten as (x-py)^2 - (py)^2 + y^2, indicating a method of completing the square.
  • Others argue that the original expression is already equivalent to the proposed form and question the necessity of completing the square.
  • A participant describes completing the square in a geometric context, suggesting that visualizing the process can aid understanding.
  • Some participants express confusion about the steps involved and request clearer explanations and examples.
  • One participant describes their own approach to rearranging the equation and completing the square, leading to a derived form of the expression.
  • Another participant seeks clarification on subsequent steps, specifically regarding the emergence of the term (1-p)^2.
  • Some participants note the importance of understanding the distributive property in simplifying expressions involving common terms.

Areas of Agreement / Disagreement

Participants generally express uncertainty about the steps involved in completing the square, with multiple competing views on the necessity and clarity of the process. The discussion remains unresolved regarding the best approach to complete the square for the given expression.

Contextual Notes

Some participants mention specific terms and assumptions, such as the role of 'p' as a constant or variable, which may not have been clearly defined in the initial discussion.

tuanle007
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x^2 + y^2 - 2pxy = (x-py)^2 - (py)^2 + y^2

can someone please help me with this. i totally forgot how to do completing the square.
 
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tuanle007 said:
x^2 + y^2 - 2pxy = (x-py)^2 - (py)^2 + y^2

can someone please help me with this. i totally forgot how to do completing the square.

you yust did complete the square. you only have a term of the form (x-A)^2 left and terms that do not depend on x.
 
Last edited:
When you perform the right-hand side multiplication and simplification, the right-hand side becomes x^2 + y^2 - 2pxy, which is the same as the left-hand side. Why do you want to complete the squares on this?
 
yeah...thats my teacher doing that..
i just want to refresh my memory on how that work..
i still don't understand...can you clearly described the steps
 
tuanle007 - Completing the square is based on starting with a rectangle and rearaanging part of it to form the sides of a square, and arithmetically & geometrically adding the missing portion in order to COMPLETE the square.

Example: x*x + b*x, where b is a constant, x is a variable. The expression can represent a rectangle with x and x+b. Notice carefully that the expression can be factored as x(x + b) = x*x + b*x.

If you draw a picture, you can see this and more in a clearer manner. Assume the height is x, and the length from left to right is x+b. Go to the middle of the 'b' portion starting at the top and cut this part in half, cutting vertically. Take the piece off, and reposition it on the left side of the entire figure so that you now have the left side of the square being x+b/2, and the top side of the square being also x+b/2. Notice now that a piece is missing at the bottom right. What is it and how much is it? It is (b/2)*(b/2).

Hopefully you can follow and understand that; the rest is up to you. How far can you go now?
 
tuanle007 said:
yeah...thats my teacher doing that..
i just want to refresh my memory on how that work..
i still don't understand...can you clearly described the steps

I tried uploading a figure about my previous post; not sure it will work.
 

Attachments

i think i kinda remember it now..
what i did was I put everything on the left side as X and everything on the right as Y
so i have
x^2 - 2pxy = -y^2
then I complete the X on the left side
(x^2 - 2pxy) + (2py/2)^2 = -y^2 + (2py/2)^2

the left side factored out to be

(x-py)^2 = -y^2 + py^2
so then move everything to the left we get

(x-py)^2 + y^2 - py^2
 
ok, i understand the completing the square part,
but can someone explain how the next step is..
where does the (1-p)^2 comes about and things like that
thanks
 

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tuanle007 said:
ok, i understand the completing the square part,
but can someone explain how the next step is..
where does the (1-p)^2 comes about and things like that
thanks

That expression was not part of your original equation. Also, 'p' was not specifically called a constant nor variable.
 
  • #10
tuanle007 said:
i think i kinda remember it now..
what i did was I put everything on the left side as X and everything on the right as Y
so i have
x^2 - 2pxy = -y^2
then I complete the X on the left side
(x^2 - 2pxy) + (2py/2)^2 = -y^2 + (2py/2)^2

the left side factored out to be

(x-py)^2 = -y^2 + py^2
so then move everything to the left we get

(x-py)^2 + y^2 - py^2

tuanle007 said:
ok, i understand the completing the square part,
but can someone explain how the next step is..
where does the (1-p)^2 comes about and things like that
thanks
That's the easy part! Long before you learned to "complete the square" you learned that "ax+ ay= a(x+ y)", the "distributive law" in technical terms. There is a y^2 term in both y^2 and -py^2 so : y^2- py^2= y^2(1)- y^2(p)= y^2(1- p).
 
  • #11
yeah..i figured it out after i sat there for 2 hours..
you guys are the best..
 

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