Solving for Smallest Value: Completing the Square for x and y

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The discussion centers on finding the smallest value of the expression x^4 + 2x^2 + y^4 - 2y^2 + 3 by completing the square. The expression is simplified to (x^2 + 1)^2 + (y^2 - 1)^2 + 1. Participants clarify that the smallest value occurs when x=0 and y=1, resulting in a minimum value of 2. There is a misunderstanding regarding the calculation, particularly about the constant term adjustment, which is corrected from 1 to 2. The conversation emphasizes the importance of understanding the behavior of squared terms in determining minimum values for real x and y.
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I never usually have a problem with it but this threw me because of the y terms.

x^4 + 2x^2 + y^4 -2y^2 + 3

i reduced it to

(x^2 + 1)^2 -1 + (y^2 -1)^2 -1 +3
=(x^2 + 1)^2 + (y^2 -1)^2 +1

but the question asks for "the smallest value (for real x and y)"
although judging by the answer to the last question, its the smallest y value.
how do i determin the value?
 
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I didn't quite get what you meant by smallest value. But I'm assuming as follows,

The least value of the function (x^2+1)^2+(Y^2-1)^2+1 for which x,y are real, which can be found out by putting x=0,y=1. So the answer comes out to be 2.
 
where did you get 0 and 1 from? trial and error?
you are correct though,
thank you

edit: don't worry i can see now
 
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Yes, through trial and error. Maybe someone else might be able to help you with the actual method.
 
Hypochondriac said:
I never usually have a problem with it but this threw me because of the y terms.

x^4 + 2x^2 + y^4 -2y^2 + 3

i reduced it to

(x^2 + 1)^2 -1 + (y^2 -1)^2 -1 +3
=(x^2 + 1)^2 + (y^2 -1)^2 +1

but the question asks for "the smallest value (for real x and y)"
although judging by the answer to the last question, its the smallest y value.
how do i determin the value?

I would interpret "the smallest value (for real x and y)" as meaning the smallest value for the expression. Since x^2+ 1 and y^2- 1 both squared (and we are told that x and y must be real) neither can be negative. The total will have the smallest value which each of those has its smallest value. What is the smallest that x^2+ 1 can be? What is the smallest y^2- 1 can be?

Oh, by the way -1+ 3= 2, not 1!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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