Quadratic System: x^2+4y^2-4x-8y+4=0, x^2+4y-4=0

S

SAR

New member
Joined
May 5, 2008
Messages
5
I need help on this system:

x[sup:17n2vjzt]2[/sup:17n2vjzt] + 4y[sup:17n2vjzt]2[/sup:17n2vjzt] - 4x -8y +4 = 0
x[sup:17n2vjzt]2[/sup:17n2vjzt] + 4y - 4 = 0

I tried substitution and linear combination, but I couldn't solve it. If someone could just tell me what to do, that would be great, my book isn't helping. (you know how books only have examples of easy problems.)
Please reply as quickly as possible.
 
If you factor, it may make it easier.

\(\displaystyle x^{2}+4y^{2}-4x-8y+4=x(x-4)+4(y-1)^{2}\)

\(\displaystyle x^{2}+4y-4=x^{2}+4(y-1)\)
 
SAR said:
I need help on this system:

x[sup:1m9uxovg]2[/sup:1m9uxovg] + 4y[sup:1m9uxovg]2[/sup:1m9uxovg] - 4x -8y +4 = 0
x[sup:1m9uxovg]2[/sup:1m9uxovg] + 4y - 4 = 0
Click to expand...
I will guess that the instructions (not included within your post) were something along the lines of "solve the following system of non-linear equations".

SAR said:
I tried substitution and linear combination, but I couldn't solve it. If someone could just tell me what to do....
Click to expand...
What to do? Um... "Try substitution or linear combinations"...? :shock:

Since you've tried what would normally be recommended, it might help if you showed your work, so we could see where you might be going wrong. Please be complete. Thank you! :D

Eliz.
 
Sorry about forgetting to post the instructions, but you were right. "find the points of intersection" were the exact instructions. I know it probably wasn't smart, but I erased a lot of my work when I tried a different method, so if someone could solve this system and show what they're doing, I guess that would be more helpful.
Thanks
 
Okay, I tried it again and this is what I have:

x(x-4) + 4(y-1)[sup:39tthhuw]2[/sup:39tthhuw]=0
x[sup:39tthhuw]2[/sup:39tthhuw] + 4(y-1) =0 , (y-1)= -x[sup:39tthhuw]2[/sup:39tthhuw]/4

x(x-4) + 4(-x^2/4)[sup:39tthhuw]2[/sup:39tthhuw]=0
x(x-4) +4(x[sup:39tthhuw]4[/sup:39tthhuw]/16)=0 , x(x-4) + (x[sup:39tthhuw]4[/sup:39tthhuw]/4)

4x(x-4) + x[sup:39tthhuw]4[/sup:39tthhuw]=0
4x[sup:39tthhuw]2[/sup:39tthhuw] - 16x + x[sup:39tthhuw]4[/sup:39tthhuw]=0
x(x[sup:39tthhuw]3[/sup:39tthhuw] + 4x -16) = 0
This is where I really get stuck, but I still don't know if I'm doing the rest right.
 
y=(4-x)/4

\(\displaystyle x^2+4(\frac{4-x^2}{4})^2-4x-8(\frac{4-x^2}{4})+4=0\)

\(\displaystyle x^2+4(\frac{16-8x^2+x^4)}{16}-4x-8-2x^2+4=0\)

\(\displaystyle 4x^2+16-8x^2+x^4-16x-32+8x^2+16=0\)

This should lead to x(x^3 + 4x -16) = 0
Now use synthetic or long division on the trinomial. Try x-2 for starters.

CORRECTED
 
Thank you all so much! I finally solved it.
Thanks again,
SAR
 
Loren: Check your multiplication: -8*((4-x^2)/4) does not yield -2x^2 but +2x^2. The correct answer to the problem is x=2, y=0, which checks for the original set of equations, but not for x*(x^3-12x-16)=0, which was your preliminary result.
 
x2 + 4y2 - 4x -8y +4 = 0
x2 + 4y - 4 = 0
Click to expand...

Graphically, this system of equations represents the intersection of an ellipse and a parabola. There are two points of intersection (solutions): (0,1) and (2,0).
 
Hi:

Including complex solutions, they are (0, 1), (2, 0), (-1 + sqrt(7)i, 0.5[5 + sqrt(7)i]), (-[1 + sqrt(7)i], 0.5[5 - sqrt(7)i]). :D

Regards,

Rich B.
rmath4u2@aol.com
 
Thank you everybody for your help. My teacher and I were both getting stuck on this one, but now I've figured it out (and with the right answers!)
-SAR
 
You must log in or register to reply here.
Top