Finding All Complex Solutions, Both Real and Imaginary: 8x^3 +27 = 0

[gretacap]

"Solve the polynomial equation. Find all complex solutions, both real and imaginary."

8x^3 +27 = 0

 

[pka]

"Solve the polynomial equation. Find all complex solutions, both real and imaginary." 8x^3 +27 = 0

\(\displaystyle x^3=\frac{-27}{8}=\frac{27}{8}\exp(\pi\bf{i})\)
Thus \(\displaystyle x=\frac{3}{2}\left(\frac{\pi(1+2k)\bf{i}}{3}\right),~k=0,1,2\)

 

[stapel]

"Solve the polynomial equation. Find all complex solutions, both real and imaginary."

8x^3 +27 = 0

Okay. They've given you a sum of cubes, along with a formula for factoring sums of cubes. (here) Where are you stuck in plugging this information into that formula?

Please be complete. Thank you!

 

[Deleted member 4993]

"Solve the polynomial equation. Find all complex solutions, both real and imaginary."

8x^3 +27 = 0

Since it is a cubic equation,you know that it has at least one real (root) solution.

What is the "real root" of this equation?

 

[HallsofIvy]

Personally, although converting to a complex exponential is the best way to solve hard equations of this type, for something a simple as this, I would just write the equation as \(\displaystyle x^3= -\frac{27}{8}\) so that one obvious solution is \(\displaystyle x= -\sqrt[3]{\frac{27}{8}}= -\frac{3}{2}\). Now divide \(\displaystyle 8x^3+ 27\) by \(\displaystyle 2x+ 3\) to get \(\displaystyle 4x^2- 6x+ 9\) and complete the square or use the quadratic formula to solve that.

 

[Deleted member 4993]

Personally, although converting to a complex exponential is the best way to solve hard equations of this type, for something a simple as this, I would just write the equation as \(\displaystyle x^3= -\frac{27}{8}\) so that one obvious solution is \(\displaystyle x= -\sqrt[3]{\frac{27}{8}}= -\frac{3}{2}\). Now divide \(\displaystyle 8x^3+ 27\) by \(\displaystyle 2x+ 3\) to get \(\displaystyle 4x^2- 6x+ 9\) and complete the square or use the quadratic formula to solve that.

Another way:

Use standard factorization (I think suggested by Stapel before):

a³ + b³ = (a + b)(a² - ab + b²)

 

[stapel]

Another way:

Use standard factorization (I think suggested by Stapel before):

a³ + b³ = (a + b)(a² - ab + b²)

Exactly. By applying the formula they gave the student, s/he would obtain the linear factor (and thus the real solution), and then could apply the Quadratic Formula to what remains (and thus obtain the two complex solutions).