How do I find the non-integer roots of the equation #x^3=3x^2+6x+2# ?
1 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^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
#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)#