# How do you find all solutions to #x^5+243=0#?

##### 2 Answers

#### Explanation:

We will make use of the fact that

Due to the periodic nature of

Note that as

Solutions in

#x = -3#

#x = (3(1+sqrt(5)))/4 +- (3sqrt(10 - 2sqrt(5)))/4i#

#x = (3(1-sqrt(5)))/4 +- (3sqrt(10 + 2sqrt(5)))/4i#

#### Explanation:

Here's an alternative approach to find the solutions in

**Fifth roots of #1#**

First let us find the Complex fifth roots of

#x^5-1 = (x-1)(x^4+x^3+x^2+x+1)#

#1/(x^2)(x^4+x^3+x^2+x+1) = x^2+x+1+1/x+1/x^2#

#color(white)(1/(x^2)(x^4+x^3+x^2+x+1)) = (x+1/x)^2+(x+1/x)-1#

Using the quadratic formula we find zeros:

#x+1/x = -1/2+-sqrt(5)/2#

Hence:

#x^2+(1/2+-sqrt(5)/2)x+1 = 0#

**Case #bb(x^2+(1/2+sqrt(5)/2)x+1 = 0)#**

Using the quadratic formula:

#x = (-(1/2+sqrt(5)/2) +- sqrt((1/2+sqrt(5)/2)^2-4))/2#

#color(white)(x) = (-(1+sqrt(5)) +- sqrt((1+sqrt(5))^2-16))/4#

#color(white)(x) = (-(1+sqrt(5)) +- sqrt(2sqrt(5)-10))/4#

#color(white)(x) = -(1+sqrt(5))/4 +- sqrt(10 - 2sqrt(5))/4i#

**Case #bb(x^2+(1/2-sqrt(5)/2)x+1 = 0)#**

Similarly (or simply reversing the signs of the coefficients of

#x = -(1-sqrt(5))/4 +- sqrt(10 + 2sqrt(5))/4i#

**Solve #bb(x^5+243) = 0#**

Since

So the solutions are:

#x = -3#

#x = (3(1+sqrt(5)))/4 +- (3sqrt(10 - 2sqrt(5)))/4i#

#x = (3(1-sqrt(5)))/4 +- (3sqrt(10 + 2sqrt(5)))/4i#

**Footnote**

Since we were able to construct the