# Solving the value

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• February 14th 2013, 05:20 AM
RuyHayabusa
Solving the value
If x^2 + y^2 = 1 then how do you find 2(x^4)+3(x^2*y^2)+(y^4)+(y*2)
• February 14th 2013, 06:18 AM
a tutor
Re: Solving the value
Replace y^2 with 1-x^2 and simplify.
• February 14th 2013, 07:35 AM
Soroban
Re: Solving the value
Hello, RuyHayabusa!

Quote:

$\text{If }x^2 + y^2 \:=\:1,\,\text{ find: }\,2x^4+3x^2y^2+y^4+y^2$

$\text{We have: }\:(2x^4 + 3x^2y^2 + y^4) + y^2 \;\;=\;\;(2x^2+y^2)\underbrace{(x^2+y^2)}_{ \text{This is 1}} + y^2$

. . . . . . . . $=\;\;(2x^2+y^2)\!\cdot\!1 + y^2 \;\;=\;\;2x^2 + y^2 + y^2 \;\;=\;\;2x^2 + 2y^2$

. . . . . . . . $=\;\;2\underbrace{(x^2+y^2)}_{\text{This is 1}} \;\;=\;\;2\!\cdot\!1 \;\;=\;\;2$
• February 15th 2013, 02:41 AM
RuyHayabusa
Re: Solving the value
Thanks for your help both.