# Find complex values of x = root (3)(343) ?

Mar 10, 2018

$x = 7$ and $x = \frac{- 7 \pm 7 \sqrt{3} i}{2}$

#### Explanation:

Assuming you mean the complex roots of the equation:

${x}^{3} = 343$

We can find the one real root by taking the third root of both sides:

$\sqrt[3]{{x}^{3}} = \sqrt[3]{343}$

$x = 7$

We know that $\left(x - 7\right)$ must be a factor since $x = 7$ is a root. If we bring everything to one side, we can factor using polynomial long division:

${x}^{3} - 343 = 0$

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

We know when $\left(x - 7\right)$ equals zero, but we can find the remaining roots by solving for when the quadratic factor equals zero. This can be done with the quadratic formula:

${x}^{2} + 7 x + 49 = 0$

$x = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \cdot 1 \cdot 49}}{2}$

$\implies \frac{- 7 \pm \sqrt{49 - 196}}{2}$

$\implies \frac{- 7 \pm \sqrt{- 147}}{2}$

$\implies \frac{- 7 \pm i \sqrt{49 \cdot 3}}{2}$

$\implies \frac{- 7 \pm 7 \sqrt{3} i}{2}$

This means that the complex solutions to the equation ${x}^{3} - 343 = 0$ are
$x = 7$ and
$x = \frac{- 7 \pm 7 \sqrt{3} i}{2}$