Question

How do you find all the real and complex roots of `125x^3 + 343 = 0`?

Answer 1

`-7/5` and `(7+-7sqrt(3)i)/10`

Explanation:

You can determine the real root of

`125x^3+343=0`

like this

`x=root(3)(-343/125)`

The real solution is `-7/5`

But you can determine all the roots by solving the equation in complex form (remember that `e^(pii)=-1)` :

`x=root(3)((343/125)e^(pii))`

So we would have the solutions:

`root(n)(rho e^(thetapii))= root(n)(rho)xxe^(((2k+thetapi)/n)i), k in ZZ`

`x=7/5xxe^(((pi+2kpi)/3)i)`

When k is 1, we obtain the real solution -7/5.

When k is 0 or -1, we obtain

`x=7/5xxe^((+-pii)/3)=7/5xx[cos(+-pii/3)+isin(+-pii/3)]`

which is

`=7/5xx(1+-sqrt(3))/2=(7+-7sqrt(3)i)/10`

Answer 2

`x=-7/5,7/10+-(7sqrt3)/10i`

Explanation:

Factor as a sum of cubes, since

`125x^3=(5x)^3`
`343=7^3`

Sums of cubes factor as follows:

`a^3+b^3=(a+b)(a^2-ab+b^2)`

Thus,

`(5x)^3+7^3=(5x+7)((5x)^2-5x(7)+7^2)`

`0=(5x+7)(25x^2-35x+49)`

This can be split up into two equations:

`5x+7=0" "=>" "x=-7/5`

And

`25x^2-35x+49=0`

Use the quadratic equation:

`x=(35+-sqrt((-35)^2-4(25)(49)))/(2(25))`

`x=(35+-sqrt(1225-4900))/50`

`x=(35+-sqrt(-3675))/50`

`x=(35+-sqrt(3 * 5^2 * 7^2)i)/50`

`x=(35+-35sqrt3i)/50`

`x=7/10+-(7sqrt3)/10i`