Thread: Quadratic-Quadratic Systems Algebraically Question

Solve Each System:

x² + y² = 20
y = x²

This might be a dumb question question, but 20 isn't a square number, so how do i find the radius to graph a circle? Please help!

Originally Posted by Vince604

Solve Each System:

x² + y² = 20
y = x²

This might be a dumb question question, but 20 isn't a square number, so how do i find the radius to graph a circle? Please help!

It has

No, it doesn't. It has radius sqrt{20}= sqrt{4(5)}= 2 sqrt{5} which is about 4.472.

(Been hitting the eggnog a little early?)

Hello, Vince604!

Solve:
. . x² + y² = 20 . [1]
. . . . y = x² . . . [2]

Substitute [2] into [1]: .y + y² = 20 . → . y² + y - 20 = 0

. . . (y - 4)(y + 5) .= .0 . → . y = 4, -5


Substitute into [2]:

. . 4 = x² . → . x = +2

. .-5 = x² . No real roots


Therefore: .(x, y) .= .(+2, 4)

Originally Posted by Soroban

Hello, Vince604!


Substitute [2] into [1]: .y + y² = 20 . → . y² + y - 20 = 0

. . . (y - 4)(y + 5) .= .0 . → . y = 4, -5


Substitute into [2]:

. . 4 = x² . → . x = +2

. .-5 = x² . No real roots


Therefore: .(x, y) .= .(+2, 4)

Thanks, it makes a lot of sense now.