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

Mar 2, 2016

$- \frac{7}{5}$ and $\frac{7 \pm 7 \sqrt{3} i}{10}$

Explanation:

You can determine the real root of

$125 {x}^{3} + 343 = 0$

like this

$x = \sqrt[3]{- \frac{343}{125}}$

The real solution is $- \frac{7}{5}$

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

$x = \sqrt[3]{\left(\frac{343}{125}\right) {e}^{\pi i}}$

So we would have the solutions:

$\sqrt[n]{\rho {e}^{\theta \pi i}} = \sqrt[n]{\rho} \times {e}^{\left(\frac{2 k + \theta \pi}{n}\right) i} , k \in \mathbb{Z}$

$x = \frac{7}{5} \times {e}^{\left(\frac{\pi + 2 k \pi}{3}\right) i}$

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

When k is 0 or -1, we obtain

$x = \frac{7}{5} \times {e}^{\frac{\pm \pi i}{3}} = \frac{7}{5} \times \left[\cos \left(\pm \pi \frac{i}{3}\right) + i \sin \left(\pm \pi \frac{i}{3}\right)\right]$

which is

$= \frac{7}{5} \times \frac{1 \pm \sqrt{3}}{2} = \frac{7 \pm 7 \sqrt{3} i}{10}$

Mar 3, 2016

$x = - \frac{7}{5} , \frac{7}{10} \pm \frac{7 \sqrt{3}}{10} i$

Explanation:

Factor as a sum of cubes, since

$125 {x}^{3} = {\left(5 x\right)}^{3}$
$343 = {7}^{3}$

Sums of cubes factor as follows:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Thus,

${\left(5 x\right)}^{3} + {7}^{3} = \left(5 x + 7\right) \left({\left(5 x\right)}^{2} - 5 x \left(7\right) + {7}^{2}\right)$

$0 = \left(5 x + 7\right) \left(25 {x}^{2} - 35 x + 49\right)$

This can be split up into two equations:

$5 x + 7 = 0 \text{ "=>" } x = - \frac{7}{5}$

And

$25 {x}^{2} - 35 x + 49 = 0$

$x = \frac{35 \pm \sqrt{{\left(- 35\right)}^{2} - 4 \left(25\right) \left(49\right)}}{2 \left(25\right)}$

$x = \frac{35 \pm \sqrt{1225 - 4900}}{50}$

$x = \frac{35 \pm \sqrt{- 3675}}{50}$

$x = \frac{35 \pm \sqrt{3 \cdot {5}^{2} \cdot {7}^{2}} i}{50}$

$x = \frac{35 \pm 35 \sqrt{3} i}{50}$

$x = \frac{7}{10} \pm \frac{7 \sqrt{3}}{10} i$