Solving Cubic Equation: x^3 + 27 = 0 | Quadratic Formula | Complex Solutions

[Menomena]

Homework Statement

x^3 +27 =0

Homework Equations

\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}

The Attempt at a Solution

(x +3)(x^2 -3x +9)= 0

(x +3) = 0, x = -3

(x^2 -3x +9)= 0

Here is where my problem starts with this equation:

I use the quadratic formula to get x= (3 plus/minus sqrt(9 -36)) / 2

Which comes out to, x = 3 plus minus 3i sqrt(3)/ 2

My book says the answer is $$\frac{{3}}{2} \pm \frac{3\sqrt{3}}{2}i$$

I understand the 3/2 but how did $$\frac{3i\sqrt{3}}{2}$$ become $$\frac{3\sqrt{3}}{2}i$$

Isn't the first way simplified enough?

 

[gb7nash]

I understand the 3/2 but how did $$\frac{3i\sqrt{3}}{2}$$ become $$\frac{3\sqrt{3}}{2}i$$

They're both the same thing. Generally, when you write a complex expression, you put the i at the very end.

 

[ehild]

They are the same. The solution is written in the standard form of complex numbers: u+v i.

ehild

edit: gb7nash beat me :)

 

[Menomena]

They're both the same thing. Generally, when you write a complex expression, you put the i at the very end.

They are the same. The solution is written in the standard form of complex numbers: u+v i.

ehild

edit: gb7nash beat me :)

Thank you both very much.