Simple question, -y^2 * -y^2 = ?

trip20

New member
Joined
Nov 4, 2006
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22
I'm trying to solve the system:

x+y^2=4
x^2+y^2=16

I set the first equation to x=4-y^2 then substituted it in the 2nd equation to get,

(4-y^2)^2+y^2=16 which I open up to:

(4-y^2)(4-y^2)+y^2=16 now when I try to FOIL I get stuck on the -y^2*-y^2

16-8y^2 ???? +y^2=16

is -y^2*-y^2 = y^4 ? or am I doing theis whole equation wrong?
 

skeeter

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Dec 15, 2005
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\(\displaystyle \L (4 - y^2)^2 = 16 - 8y^2 + y^4\)

but ... if I were you, I'd use the 1st equation and solve for y<sup>2</sup>, then substitute what you get for y<sup>2</sup> in the 2nd equation ... might be a 'tad' easier.
 

trip20

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Nov 4, 2006
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22
thanks skeeter,

your way does make more sense, thank you.

however i still get stuck:

y^2=4-x

x^2+4-x=16

x^2-x=12

then what ??? I know x=4 but I'm not sure how to get there for here
 

skeeter

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Dec 15, 2005
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x<sup>2</sup> - x = 12

x<sup>2</sup> - x - 12 = 0

(x - 4)(x + 3) = 0

two possible solutions
x = 4, x = -3

one at a time ... substitute each solution for x back into y<sup>2</sup> = 4 - x and solve for y ... (you'll get three solutions for y)
 

Denis

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Feb 17, 2004
Messages
1,489
trip20 said:
I'm trying to solve the system:
x + y^2 = 4 [1]
x^2 + y^2 = 16 [2]
Keep it simple, trip; subtract [1] from [2]:

x^2 - x + y^2 - y^2 = 16 - 4

x^2 - x = 12

x^2 - x - 12 = 0
 
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