SOLUTION: Solve this equation: x^3+729=0 I know this much: 1st solution is -9 (x+9)^3=0 (x+9)[x^2-9x+81] This is where I'm stuck..When I use the quadratic formula I get 9 +/-squa

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve this equation: x^3+729=0 I know this much: 1st solution is -9 (x+9)^3=0 (x+9)[x^2-9x+81] This is where I'm stuck..When I use the quadratic formula I get 9 +/-squa      Log On


   

Question 260277: Solve this equation: x^3+729=0
I know this much: 1st solution is -9
(x+9)^3=0
(x+9)[x^2-9x+81]
This is where I'm stuck..When I use the quadratic formula I get
9 +/-square root of 243 over 2
Now what? The answer is 9 +9i 'square root' of 3 over 2. How do you get that?
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Here is the original question:
x%5E3%2B729=0+
we can express this as the sum of cubes, but first write them both as powers of 3, as
x%5E3%2B9%5E3+=+0
Now, we get
%28x%2B9%29%28x%5E2-9x%2B81%29=0
next set each part =0 and solve.
x+9 = 0
x = -9
--
x^2-9x+81= 0
by quadratic, we get
x+=+%289+%2B-+sqrt%28+81-4%2A1%2A81+%29%29%2F%282%29+
which simplifies to
x+=+%289+%2B-+sqrt%2881-324%29%29%2F%282%29+
which is
x+=+%289+%2B-+sqrt%28-243+%29%29%2F%282%29+
which will give us imaginary roots as
x+=+%289+%2B-+9%2Aisqrt%283%29%29%2F2
--
we have three roots
x = -9
x = (9+9isqrt(3))/2
x = (9-9isqrt(3))/2