Help solve this equation

x^3 + x^2 - x =0


help me solve this please!
any help would be greatly appreciated!

x^3 + x^2 - x =0
x(x^2 + x -1) = 0
for the equation x^2 + x -1
x = [-b±√(b^2-ac)]/2a
x = [-1±√(1^2-4X1X-1)]/2X1
x = [-1±√(1+4)]/2
x = [-1±√(5)]/2

x^3 + x^2 - x =0
x(x^2 + x -1) = 0
x = 0 or x^2 + x -1 = 0
x = 0 or x = (-1+√5)/2 or x = (-1-√5)/2

You have to factor it.

x^3 + x^2 - x = 0
x(x^2 + x - 1) = 0. Since the factors are prime polynomials, set each to 0.

x = 0 (one solution)
x^2 + x - 1 = 0

x = (-1 +- sqrt(1^2 - 4(1)(-1)))/2(1)
x = (-1 +- sqrt(1 +4))/2
x = (-1 +- sqrt(5))/2
These are the exact solutions, (-1 + sqrt(5))/2 and (-1 - sqrt(5))/2, you could use a calculator to get approximate solutions, if necessary.

Take x out:
x(x^2 + x -1) = 0

Put expression in the brackets into the quadratic formula:

x = -1 +/- sqrt(1 - 4(1)(-1)) / 2
= -1 +/- sqrt(5) / 2
so x = (-1 + sqrt(5)) / 2
x = (-1 - sqrt(5)) / 2
x = 0 (because of the x we took out the front earlier on, which was really a factor as (x + 0))

Hope this helps!


PS. If you don't know the quadratic formula, heres what it is:

if you let y = ax^2 + bx + c

solutions for x = [ -b +/i sqrt(b^2 - 4ac) ] / 2a

:)

PPS: My answer is correct, just checked it on MATLAB :D Graph it yourself and take a look!

take x common
x(x^2 + x -1) = 0
therefore one of the solutions is x=0
now another solution is when x^2+ x -1=0
its a quadratic , after solving u get x= -1(+-)sqrt 5 /2
therefore your answers are x=0
x=-1+sqrt5/2
x= -1-sqrt5/2

factor bro.
x(x^2 + x -1) = 0
quadratic formula.
x = 0, -1.618, 0.62

x^3 + x^2 - x =0

x(x^2 + x -1) = 0 =>

a) x= 0

b) x^2 + x - 1 =0 => x(x+1) = 1 which is impossible

so x=0

x=0 maybe? thats all i can think of which would add up right