What are the roots of #x^3-8x-3=0#?

1 Answer
Mar 24, 2017

Answer:

Three roots of #x^3-8x-3=0# are #3#, #(-3+sqrt5)/2# and #(-3-sqrt5)/2#

Explanation:

in an equation of type #x^n+a_1x^(n-1)+a_2x^(n-2)+.........+a_n#, the roots of the equation are factors of #a_n#.

Here, we have the equation as #x^3-8x-3=0# and hence roots are factors of #3# i.e. #+_1# and #+-3#.

It is apparent that #x=3# is a root as it satisfies the equation

#3^3-8xx3-3=27-24-3=0# and hence

#x-3# is a factor of #x^3-8x-3# and dividing latter by former

#x^2(x-3)+3x(x-3)+1(x-3)=(x-3)(x^2+3x+1)#

As such we have #(x-3)(x^2+3x+1)=0#

And as #x^2+3x+1# cannot be factorized as rational factors, using quadratic formula roots are

#(-3+-sqrt(3^2-4xx1xx1))/2=(-3+-sqrt5)/2#

and hence three roots of #x^3-8x-3=0# are #3#, #(-3+sqrt5)/2# and #(-3-sqrt5)/2#