Thread: Nonlinear system, manually computing imaginary solutions

1. Nonlinear system, manually computing imaginary solutions

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

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 $\displaystyle 25x^2-36x-108$ 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 $\displaystyle \frac{36\pm \sqrt{12096}}{50}$ Which does not make much sense.

2. Re: Nonlinear system, manually computing imaginary solutions

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

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

3. Re: Nonlinear system, manually computing imaginary solutions

Hello, Greymalkin!

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

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

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

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

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

4. Re: Nonlinear system, manually computing imaginary solutions

Hello agaibn, Greymalkin!

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

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

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

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

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

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

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

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

5. Re: Nonlinear system, manually computing imaginary solutions

Thanks again Soroban!

6. Re: Nonlinear system, manually computing imaginary solutions

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

Second system has two real and two complex roots: