# Solving Quadratic Systems of Equations

• May 10th 2008, 01:52 PM
mathgeek777
Solving Quadratic Systems of Equations
I have this system:

$x^2+2y^2-4x+6y-5=0$
$-x+y-4=0$

And I have to find the intersecting points of the two equations by using substitution and only substitution.

I solved the bottom equation for y getting $y=x+4$. From there, this is how I progressed:

$x^2+2(x+4)^2-4x+6(x+4)-5=0$
$x^2+2(x^2+8x+16)-4x+6x+24-5=0$
$x^2+2x^2+16x+32-4x+6x+24-5=0$

Combining like terms:

$3x^2+18x+51=0$

Pulling out a 3 gives me:

$3(x^2+6x+17)=0$

From here, I'm not sure what to do. Am I on the right track or am I doing this completely wrong? If the latter, please explain to me where I made an error so it doesn't happen again.

Thank you.
• May 10th 2008, 02:06 PM
Moo
Hello,

Quote:

Originally Posted by mathgeek777
I have this system:

$x^2+2y^2-4x+6y-5=0$
$-x+y-4=0$

And I have to find the intersecting points of the two equations by using substitution and only substitution.

I solved the bottom equation for y getting $y=x-4$. From there, this is how I progressed:

$x^2+2(x+4)^2-4x+6(x{\color{red}+}4)-5=-$
$x^2+2(x^2+8x+16)-4x+6x+24-5=0$
$x^2+2x^2+16x+32-4x+6x+24-5=0$

Combining like terms:

$3x^2+18x+51=0$

Pulling out a 3 gives me:

$3(x^2+6x+17)=0$

From here, I'm not sure what to do. Am I on the right track or am I doing this completely wrong? If the latter, please explain to me where I made an error so it doesn't happen again.

Thank you.

Little typo in red, not changing what follows.

Well, I don't see any mistake that far.

I've substituted y, and I also get complex solutions :-)
• May 10th 2008, 02:11 PM
mathgeek777
Since I couldn't find any obvious real solution, I put it down as no real solutions, checked the back of the book, and it is indeed no real solutions.

Thanks for the help.