Question

How do you solve `x^ { 5} - 2= 0`?

Answer 1

`x approx {1.14870, 0.35497+1.09248i, -0.92932+0.67519i, 0.35497-1.09248i, -0.92932-0.67519i}`

Explanation:

`x^5 -2=0`

`x^5 =2`

Since the equation is a polynomial of degree 5 with real coefficients we expect to find 5 real/complex roots.

Since the complex roots will occur in conjugate pairs, there must be either 1 or 3 real roots.

The obvious real root is: `x=root5(2) approx 1.14870`

From De Moivre's formula* we can deduce for this example:

`x=2^(1/5){cos((2npi)/5) + i sin((2npi)/5)}` for `n in [0,4]`

`n=0` yields the real root `2^(1/5) approx 1.14870`

`n=1` yields `x approx 0.35497+1.09248i`

`n=2` yields `x approx -0.92932+0.67519i`

Since `n=1 and n=2` yields roots that are not a conjugate pair, the equation must have 4 complex roots. The remaining 2 complex roots must form conjugate pairs with the complex roots above.

Hence, `x approx {0.35497-1.09248i, -0.92932-0.67519i}` must be the 2 remaining complex roots.

So the complete solution of the equation is the following 5 roots:

`x approx {1.14870, 0.35497+1.09248i, -0.92932+0.67519i, 0.35497-1.09248i, -0.92932-0.67519i}`

*For further informaion on De Moivre's formula, see https://en.wikipedia.org/wiki/De_Moivre%27s_formula#Roots_of_complex_numbers