Completing The Square

  • #1
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?
 

Answers and Replies

  • #2
60
0
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.
 
  • #3
where did you get 0 and 1 from? trial and error?
you are correct though,
thank you

edit: dont worry i can see now
 
Last edited:
  • #4
60
0
Yes, through trial and error. Maybe someone else might be able to help you with the actual method.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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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?
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!
 

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