Question

Complete the square:
i) x^4 +2x^2 +y^4 - 2y^2 +3

Answer 1

The answer is supposed to be 2

Answer 2

for simplicity,
say x^2 = m,
y^2 = n

Answer 3

Do you have any idea on applications of completing the squares?

Answer 4

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

m^2 + 2m + n^2 - 2n + 3

Answer 5

Applications? Not really. I've only ever answered questions with max powers of 2. I think ganeshie8's solution looks like it might work

Answer 6

Yes. That will really work! Haha. Do you know that next steps?

Answer 7

I've got (m+1)^2-1 + (n-1)^2 -1 +3. So then (m+1)^2 + (n-1)^2 +1

Answer 8

good job !

Answer 9

The answer is supposed to be 2

??? dint get this part

Answer 10

I then have to find the smallest value apparently (for real x and y)

Answer 11

you wanto find smallest value of function ?

Answer 12

yes

Answer 13

let me think lol it looks tough for me :|

Answer 14

Ok, yeh it is tough

Answer 15

@experimentX @hartnn

Answer 16

question is :
find minimum value of x^4 +2x^2 +y^4 - 2y^2 +3

Answer 17

did you complete the square?

Answer 18

Yes, we got (m+1)^2 + (n-1)^2 +1 where m=x^2 and n=y^2

Answer 19

there are two parts of the problem.
x^4 +2x^2 +y^4 - 2y^2 +3
--------- this part's is always positive. so minimum value is zero for x=0
the toehr part y^4 - 2y^2 +3 complete square (y^2 - 1)^2 + 2
the you get the minimum value by minimizing (y^2 - 1)^2 ... that is y=1

Answer 20

so the minimum value you get is when x=0 and y=1, that is 2

Answer 21

Ah, I get it! Thank you!

Answer 22

thanks @experimentX it was awesome !