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

2 Answers
Mar 2, 2016

Answer:

#-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#

Mar 3, 2016

Answer:

#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#