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

1 Answer
Mar 10, 2018

x=7 and x=(-7+-7sqrt(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:

root(3)(x^3)=root(3)(343)

x=7

We know that (x-7) 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

(x-7)(x^2+7x+49)=0

We know when (x-7) 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+7x+49=0

x=(-7+-sqrt(7^2-4*1*49))/2

=>(-7+-sqrt(49-196))/2

=>(-7+-sqrt(-147))/2

=>(-7+-isqrt(49*3))/2

=>(-7+-7sqrt(3)i)/2

This means that the complex solutions to the equation x^3-343=0 are
x=7 and
x=(-7+-7sqrt(3)i)/2