finding roots

[tsh44]

Hi I need help solving this problem:

z^3+ 8 = 0. I need to fidn the roots of the equation. I appreciate anyhelp.

 

[royhaas]

Hint: $\displaystyle z^3+8 = (z+2)(z^2-2z+4)$.

 

[tsh44]

okay one root is z= -2 but how can I faind the others if i can not factor the other part?

 

[Gene]

Don't you know the quadratic equation solution for ax²+bx+c=
(-b+sqrt(b^2-4ac)/2a

 

[tsh44]

Would 1+/- 2i be correct?

 

[mad_mathematician]

z^3 + 8 = 0
z^3 = -8
z= -2

thats the real root, and im not sure how you'd find the complex roots.

 

[pka]

Because $\displaystyle z^3 = - 8$ z is one of the three cube roots of –8.

Because $\displaystyle - 8 = 2^3 (\cos (\pi ) + i\sin (\pi )) = 2^3 e^{i\pi }$, the three roots are:
\(\displaystyle \begin{array}{l}
2(\cos (\pi /3) + i\sin (\pi /3)) = 2e^{i\pi /3} \\
2(\cos (\pi ) + i\sin (\pi )) = 2e^{i\pi } \\
2(\cos ( - \pi /3) + i\sin ( - \pi /3)) = 2e^{ - i\pi /3} \\
\end{array}\).

 

[galactus]

Just solve the quadratic in the factorization to find the two complex roots.

You can use the quadratic formula.

\(\displaystyle \L\\z^{2}-2z+4=0\)

\(\displaystyle \L\\\frac{-(-2)+\sqrt{(-2)^{2}-4(1)(4)}}{2(1)}=1+\sqrt{3}i\)

\(\displaystyle \L\\\frac{-(-2)-\sqrt{(-2)^{2}-4(1)(4)}}{2(1)}=1-\sqrt{3}i\)