How do I find the non-integer roots of the equation x^3=3x^2+6x+2x3=3x2+6x+2 ?
1 Answer
The non-integer roots are:
x = 2+-sqrt(6)x=2±√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 = 0x3−3x2−6x−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 = 0−1−3+6−2=0
Hence we can tell that
x^3-3x^2-6x-2 = (x+1)(x^2-4x-2)x3−3x2−6x−2=(x+1)(x2−4x−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)a2−b2=(a−b)(a+b)
with
x^2-4x-2 = x^2-4x+4-6x2−4x−2=x2−4x+4−6
color(white)(x^2-4x-2) = (x-2)^2-(sqrt(6))^2x2−4x−2=(x−2)2−(√6)2
color(white)(x^2-4x-2) = ((x-2)-sqrt(6))((x-2)+sqrt(6))x2−4x−2=((x−2)−√6)((x−2)+√6)
color(white)(x^2-4x-2) = (x-2-sqrt(6))(x-2+sqrt(6))x2−4x−2=(x−2−√6)(x−2+√6)
Hence zeros:
x = 2+-sqrt(6)x=2±√6
