Nonlinear system, manually computing imaginary solutions

**Greymalkin** · August 27th 2012, 03:46 PM

$x^2+y^2=4$
$16x^2+9y^2=144$

The answer is not as important as how you arrived there, wolfram alpha can compute these. I would like to know how to solve manually.

I keep getting the equation $25x^2-36x-108$ from substitution(solved for formula 1, subbed into 2),
elimination yields nothing because there are no real roots, the graph is a circle within an ellipse.
Using the quadratic formula with the above equation gives me much different answers than wolfram.
I get $\frac{36\pm \sqrt{12096}}{50}$ Which does not make much sense.

**Greymalkin** · August 27th 2012, 04:56 PM

Another system I cannot solve manually is:
$5y^2-x^2=1$
$xy=2$

Due to the same reasons, I think I am forcing myself into extraneous solutions somehow :S

**Soroban** · August 27th 2012, 05:23 PM

Hello, Greymalkin!

$\begin{array}{cccc}x^2+y^2&=&4 & [1] \\ 16x^2+9y^2 &=& 144 & [2] \end{array}$

$\begin{array}{ccccccc}\text{Multiply -9}\times[1]: & \text{-}9x^2 - 9y^2 &=& \text{-}36 \\ \text{Add [2]:} & 16x^2 + 9y^2 &=& 144 \end{array}$

We have: . $7x^2 \:=\:108 \quad\Rightarrow\quad x^2 \:=\:\tfrac{108}{7} \quad\Rightarrow\quad \boxed{x \:=\:\tfrac{6}{7}\sqrt{21}}$

Substitute into [1]: . $\tfrac{108}{7} + y^2 \:=\:4 \quad\Rightarrow\quad y^2 \:=\:\text{-}\tfrac{80}{7}$

. . . . . . . . . . . . . . . $y \:=\:\pm\sqrt{\text{-}\tfrac{80}{7}} \quad\Rightarrow\quad \boxed{y\:=\:\pm\tfrac{4}{7}i\sqrt{35}}$

**Soroban** · August 27th 2012, 05:50 PM

Hello agaibn, Greymalkin!

$\begin{array}{cc}5y^2-x^2\:=\:1 & [1] \\ xy\:=\:2 & [2] \end{array}$

From [2]: . $y \,=\,\tfrac{2}{x}$

Substitute into [1]: . $5\left(\tfrac{2}{x}\right)^2 - x^2 \:=\:1 \quad\Rightarrow\quad \frac{20}{x^2} - x^2 \:=\:1$

. . $20 - x^4 \:=\:x^2 \quad\Rightarrow\quad x^4 + x^2 - 20 \:=\:0 \quad\Rightarrow\quad (x^2-4)(x^2+5) \:=\:0$

. . $\begin{Bmatrix}x^2-4 \:=\:0 & \Rightarrow & x^2 \:=\:4 & \Rightarrow & x \:=\:\pm2 \\ x^2+5 \:=\:0 & \Rightarrow & x^2 \:=\:\text{-}5 & \Rightarrow & x \:=\:\pm i\sqrt{5} \end{Bmatrix}$

If $x = \pm2$ , [2] becomes: . $(\pm2)y \:=\:2 \quad\Rightarrow\quad y \:=\:\pm1$

If $x = \pm i\sqrt{5$ , [2] becomes: . $(\pm i\sqrt{5})}y \:=\:2 \quad\Rightarrow\quad y \:=\:\mp\tfrac{2i}{5}\sqrt{5}$

We have four solutions: . $(\pm2,\,\pm1),\;\left(\pm i\sqrt{5},\:\mp\tfrac{2i}{5}\sqrt{5}\right)$

**Greymalkin** · August 27th 2012, 05:54 PM

Thanks again Soroban!

**MaxJasper** · August 27th 2012, 05:55 PM

First system has a circle & ellipse that do not interset:

Second system has two real and two complex roots:

Thread: Nonlinear system, manually computing imaginary solutions

LinkBack

Thread Tools

Display

Nonlinear system, manually computing imaginary solutions

Re: Nonlinear system, manually computing imaginary solutions

Re: Nonlinear system, manually computing imaginary solutions

Re: Nonlinear system, manually computing imaginary solutions

Re: Nonlinear system, manually computing imaginary solutions

Re: Nonlinear system, manually computing imaginary solutions

Similar Math Help Forum Discussions

System of nonlinear first-order differential equations (Matrix Riccati system)

How do I find the imaginary number solutions to this problem?

real and imaginary solutions

Solutions for imaginary numbers in a+bi

imaginary solutions

Search Tags