Roots of $x^3-2=0$...

Question

When I was solving this,$$\begin{equation}\begin{aligned} x^3-2&=0\\x^3&=2\\x&=2^{1/3}\end{aligned}\end{equation}$$,I got $x=2^{1/3}$

But this is only one root...I know there are two complex roots also...But not getting how we are getting complex roots and what are they....?

Please explain me, I am not getting.....

Thanks in advance!!

Answer 1

You have

$$x^3=2$$ i.e$$x^3-(2^\frac{1}{3})^3=0$$ i.e. $$(x-2^\frac{1}{3})(x^2+2^\frac{1}{3}x+2^\frac{2}{3})=0$$ so, $$x=2^\frac{1}{3} \text{or}$$

$$x^2+2^\frac{1}{3}x+2^\frac{2}{3}=0$$

Solve this quadratic equation on your own, you'll get what you want!

Answer 2

See $x^3=2$ is one solution but by solving the equation we get other two solutions like this. $$ x^3-2= (x)^3-(2^\frac{1}{3})^3=(x-2^\frac{1}{3})(x^2+2^\frac{1}{3}x+2^\frac{2}{3})$$.
Solving $x^2+2^\frac{1}{3}x+2^\frac{2}{3}$ you will get the other two solution which is complex because complex roots occur in conjugate and it is a quardratic equation as well.

Answer 3

Solving the Equation: $x^3 - 2 = 0$

Given Equation: $x^3 = 2$

This is a cubic equation, and we are looking for values of $x$ that satisfy the equation.

Factoring the Equation:

We can factor the equation using the difference of cubes formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. $$x^3 - 2 = (x - \sqrt[3]{2})(x^2 + x\sqrt[3]{2} + (\sqrt[3]{2})^2) = 0$$

Roots:

1. Real Root: $x = \sqrt[3]{2}$

2. Complex Roots:

For complex roots, we solve the quadratic equation obtained from factoring:

$$x^2 + x(\sqrt[3]{2}) + (\sqrt[3]{2})^2 = 0$$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ we get,

$$x \in \bigg(\frac{-1 + i\sqrt{3}}{2^{2/3}}, \frac{-1 - i\sqrt{3}}{2^{2/3}}\bigg)$$

These are the complex roots of the equation $\tag*{$\blacksquare$}$

Answer 4

It seems you could use some introduction to roots of unity.

For instance, $3$rd roots of unity are anything whose cube is $1$. It turns out that there are three of them. One of those is the familiar old number $1$. Of course, $$1^3=1.$$

The other two happen to not be real numbers. Turns out, they're $$-1/ 2+\sqrt3/2i$$ and $$-1/ 2-\sqrt3/2i.$$

As to how we find them, there's arguably a few different ways. But basically this gets you into the subject of complex numbers, as well as, rather obviously, complex analysis.

For starters, and as a point of departure, let's see Euler's formula (in full generality):

$$e^{iz}=\cos z+i\sin z.$$

This works out nicely in terms of power series and what not. I'll leave it for you to explore these ideas a little more as you progress.

One last note: the $n$th roots of unity can always be written as powers of $$e^{2\pi i/n},$$ which is thus an example of a primitive root.