# Math Help - Solve quadratic into complex numbers

1. ## Solve quadratic into complex numbers

Hi

I have this equation:
$s^2 + \sqrt{2}s + 4$

and it somehow factorises to this:

$(s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})$

I've tried using the quadratic formula but cant get it into the form with $e^{x}$

Would anyone be able to shed a little light on this?

Thanks

2. Originally Posted by eggy524
Hi

I have this equation:
$s^2 + \sqrt{2}s + 4$

and it somehow factorises to this:

$(s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})$

I've tried using the quadratic formula but cant get it into the form with $e^{x}$

Would anyone be able to shed a little light on this?

Thanks
Well it doesn't if you multiply it out you get

$(s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})=$
$\displaystyle s^2+2se^{-j\frac{\pi}{4}}+2se^{j\frac{\pi}{4}}+4$
$\displaystyle s^2+2s\left(e^{j\frac{\pi}{4}}+e^{-j\frac{\pi}{4}} \right)+4$
$\displaystyle s^2+2s(2\cos\left(\frac{\pi}{4} \right))+4$
$s^2+2\sqrt{2}s+4$

3. Originally Posted by eggy524
Hi

I have this equation:
$s^2 + \sqrt{2}s + 4$

and it somehow factorises to this:

$(s + 2e^{+j(\pi/4)})(s + 2e^{-j(\pi/4)})$

I've tried using the quadratic formula but cant get it into the form with $e^{x}$

Would anyone be able to shed a little light on this?

Thanks
When you use the quadratic formula to solve the equation you get $s=\frac{-\sqrt{2}\pm\sqrt{2- 4(1)(4)}}{2}= \frac{-\sqrt{2}\pm\sqrt{-14}}{2}$
so that $x^2+ \sqrt{2}x+ 4= \left(s+ \frac{\sqrt{2}}{2}+i\frac{\sqrt{14}}{2}\right)\lef t(s+ \frac{\sqrt{2}{2}}- i\frac{\sqrt{14}}{2}\right)$.

You can change those numbers to "polar" form but they are not anything like " $2e^{i\pi/4}$" or " $e^{-i\pi/4}$.

The modulus is correct: $\sqrt{\frac{1}{2}+ \frac{7}{2}}= \sqrt{4}= 2$
but the argument is $tan^{-1}\left(\frac{\sqrt{14}}{2}\frac{2}{\sqrt{2}}\righ t)= tan^{-1}\right(\sqrt{7})$ which is closer to 1.2 radians than to $\pi/4$.

Did you, as TheEmptySet suggests, drop a factor of "2" from the coefficient of x?

4. Thanks guys, makes more sense now!