Factorisation with 2 variables

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In summary, the conversation discusses finding the lowest possible value of a given expression without using calculus. The suggested method involves completing the squares for x and y and manipulating the expression to eliminate the xy term. The conversation also includes suggestions and hints on how to further simplify the expression. Ultimately, the final solution involves completing the square for both x and y and manipulating the expression to get the lowest possible value.
  • #1
GregA
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Is there a way to find the lowest possible value of the following expression:
[tex] x^2 + 4xy + 5y^2 - 4x - 6y +7 [/tex] without using calculus?

I've just got myself a different textbook and am working from the beginning, trying to use only the tools given to me thus far in the book (within reason).

One way to solve it is to complete the squares for x and y to get
[itex] (x-2)^2+5(y- \frac{3}{5})^2 +4xy + \frac {6}{5} [/itex] and then find the value of 4xy which keeps the sum of both squares and itself as small as possible..ie differentiate

Problem is the book has not not even touched upon calculus yet and so I suspect there is a more elementary approach that I have either missed or am not even aware of and this is bugging me :grumpy: ...The furthest I got was the factorisation above and figuring out that 4xy should be negative and its absolute value being as high as possible so that y < 0 whilst x > 2 ... but I still cannot find a basic method of finding the precise values of x and y. Anyone have any suggestions?
 
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  • #2
Try getting rid of the xy term first... by completing the square!

(Hint: instead of looking at (x-a)², try looking at (x-ay)²)
 
  • #3
Thanks for the reply Hurkyl

By completing the square to eliminate xy I get [tex] (x+2y)^2 +y^2-6y-4x+7 [/tex]...I can go further and wrap up y to get [tex] (x+2y)^2 + (y-3)^2 -4x - 2 [/tex] but I cannot see any useful clues...(apart from what I discovered from the other factorisation), nor can I see a way to force exact values upon x or y :frown:
 
  • #4
You can still complete the square to get rid of the x term. (Doing so will introduce another y term, but you already know how to get rid of that)
 
  • #5
Aha! :biggrin:

[tex] x^2 + 4xy + 5y^2 -4x -6y + 7 = [/tex]
[tex](x +2y -2)^2 + y^2 +2y +3 = [/tex]
[tex] (x +2y -2)^2 +(y+1)^2 +2 [/tex]

Thanks for steering me in the right direction Hurkyl!
 

1. What is factorisation with 2 variables?

Factorisation with 2 variables is a mathematical process that involves finding the common factors of an expression with two variables. It simplifies an expression by breaking it down into smaller, more manageable parts.

2. Why is factorisation with 2 variables important?

Factorisation with 2 variables is important because it allows us to solve equations, simplify expressions, and find the roots of polynomials. It also helps us identify patterns and relationships between variables in a given expression.

3. How do you factorise an expression with 2 variables?

To factorise an expression with 2 variables, we first look for common factors in the expression. Then, we use the distributive law to expand the expression and rearrange it to group the common factors together. Finally, we factor out the common factors to get the simplified expression.

4. Can we use the same methods for factorisation with 2 variables as we do with one variable?

Yes, the same methods used for factorisation with one variable can also be applied to factorisation with 2 variables. However, factorisation with 2 variables can be more complex and may require additional steps compared to factorisation with one variable.

5. What are some real-life applications of factorisation with 2 variables?

Factorisation with 2 variables is commonly used in engineering, physics, and economics to solve equations and simplify complex expressions. It is also used in computer programming to optimize algorithms and solve problems involving multiple variables.

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