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#