How do I find the non-integer roots of the equation #x^3=3x^2+6x+2# ?

1 Answer
Sep 5, 2016

Answer:

The non-integer roots are:

#x = 2+-sqrt(6)#

Explanation:

First subtract the right hand side of the equation from the left, to get it into standard form:

#x^3-3x^2-6x-2 = 0#

Note that if we reverse the signs of the coefficient on terms of odd degree then the sum is zero. That is:

#-1-3+6-2 = 0#

Hence we can tell that #x=-1# is a root and #(x+1)# a factor:

#x^3-3x^2-6x-2 = (x+1)(x^2-4x-2)#

We can find the zeros of the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a = (x-2)# and #b=sqrt(6)# as follows:

#x^2-4x-2 = x^2-4x+4-6#

#color(white)(x^2-4x-2) = (x-2)^2-(sqrt(6))^2#

#color(white)(x^2-4x-2) = ((x-2)-sqrt(6))((x-2)+sqrt(6))#

#color(white)(x^2-4x-2) = (x-2-sqrt(6))(x-2+sqrt(6))#

Hence zeros:

#x = 2+-sqrt(6)#