# Solving a system graphically and numerically

• December 1st 2008, 10:45 AM
LoveBeachForever
Solving a system graphically and numerically
Okay so I'm having trouble with a system involving a circle and a parabola. My book tells me to solve the system graphically and support my answer numerically. I don't know how to do this and really appreciate the help!!! :)

$x^2 + y^2 = 9$
$y=2-x^2$
• December 1st 2008, 11:09 AM
running-gag
Quote:

Originally Posted by LoveBeachForever
Okay so I'm having trouble with a system involving a circle and a parabola. My book tells me to solve the system graphically and support my answer numerically. I don't know how to do this and really appreciate the help!!! :)

$x^2 + y^2 = 9$
$y=2-x^2$

The system leads to
$x^2 = 9 - y^2$
$y=2-(9 - y^2)=-7 + y^2$

Second equation gives $y^2 - y - 7 = 0$ which is a second degree equation
• December 1st 2008, 11:51 AM
LoveBeachForever
Okay so if $0=y^2-y-7$

then I can use the quadratic and get $y=\frac{1 \pm \sqrt 29}{2}$

which is $y=3.193$ or $-2.193$
using that and plugging it back in i get:
$3.193=2-x^2$
$x^2=1.193$
$x=1.092$ so the solution is at (3.193,1.092) correct?

then for the other one,
$-2.193=2-x^2$
but $x^2=-4.193$ which leaves me with an imaginary number. So I'm assuming that it means that just the other one is the solution. Am I right?
• December 1st 2008, 12:34 PM
running-gag
Quote:

Originally Posted by LoveBeachForever
Okay so if $0=y^2-y-7$

then I can use the quadratic and get $y=\frac{1 \pm \sqrt 29}{2}$

which is $y=3.193$ or $-2.193$
using that and plugging it back in i get:
$3.193=2-x^2$
$x^2=1.193$
$x=1.092$ so the solution is at (3.193,1.092) correct?

then for the other one,
$-2.193=2-x^2$
but $x^2=-4.193$ which leaves me with an imaginary number. So I'm assuming that it means that just the other one is the solution. Am I right?

No
$3.193=2-x^2$
$x^2=-1.193$=> no solution
$-2.193=2-x^2$
but $x^2=4.193$ => 2 solutions : sqrt(4.193) and -sqrt(4.193)