Melody
Feb 15, 2014

Melody
Feb 14, 2014

#1**+1 **

There are 3 roots each 2pi/3 = 4pi/6 apart

\(\sqrt[3]{8i}=-2i\)

-2i has a anticlockwise angle of \(\frac{3\pi}{2}=\frac{9\pi}{6}\\ \)

\(\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\ \frac{5\pi}{6}-\frac{4\pi}{6}=\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\~\\ \text{so the 3 cubic roots are }\quad \\ 2e^{(\frac{\pi}{6})i},\quad 2e^{(\frac{5\pi}{6})i},\quad 2e^{(\frac{3\pi}{2})i}\\ \sqrt[3]{8i}=2(cos\frac{\pi}{6}+isin\frac{\pi}{6}),\;2(cos\frac{5\pi}{6}+isin\frac{5\pi}{6}),\;2(cos\frac{3\pi}{2}+isin\frac{3\pi}{2})\\ etc\)

checked:

https://www.wolframalpha.com/input/?i2d=true&i=Surd%5B8i%2C3%5D

LaTex:

\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\

\frac{5\pi}{6}-\frac{4\pi}{6}=\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\~\\

\text{so the 3 cubic roots are }\quad \\

2e^{(\frac{\pi}{6})i},\quad 2e^{(\frac{5\pi}{6})i},\quad 2e^{(\frac{3\pi}{2})i}\\

\sqrt[3]{8i}=2(cos\frac{\pi}{6}+isin\frac{\pi}{6}),\;2(cos\frac{5\pi}{6}+isin\frac{5\pi}{6}),\;2(cos\frac{3\pi}{2}+isin\frac{3\pi}{2})\\

etc

MelodyJan 26, 2022

#1**+1 **

Hi Julia,

These are not equations. there are not equal signs. They need y= or f(x)= or some other funtion name =

I will assume they are f(x)= and g(x)=

a) f(x)=g(x) then

-4x2+2x+5=2x2-3x+6

now solve them simultaneously.

b) plug the answers that you get into the original equation to see if they make f(x) and g(x) equal

You can also put the equations into Desmos graphing calc to see where they both cross.

For the calc call them both y=

https://www.desmos.com/calculator

** You can now show the forum what you have done in order to give yourself learning credibility here.

MelodyJan 26, 2022