SOLUTION: solve for x: 4x=-5x^2 +2x^3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: solve for x: 4x=-5x^2 +2x^3      Log On


   

Question 643589: solve for x: 4x=-5x^2 +2x^3
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
4x=-5x%5E2+%2B2x%5E3
For any polynomial equation of degree greater than 1, you want one side of the equation to be a zero. So we'll start by subtracting 4x from each side:
+0=-5x%5E2+%2B2x%5E3-4x

Before we factor, let's put the terms in order from highest exponent to lowest:
+0=+2x%5E3-5x%5E2+-4x
Now we factor. First the greatest common factor , GCF, which is x:
+0=+x%282x%5E2-5x+-4%29
This is as far as the expression will factor. From this and the Zero Product property we know that one of these factors must be zero. So
x+=+0 or 2x%5E2-5x-4+=+0
To find the solutions to the second equation, we will use the Quadratic Formula:
x+=+%28-%28-5%29%2B-sqrt%28%28-5%29%5E2-4%282%29%28-4%29%29%29%2F2%282%29
Simplifying...
x+=+%28-%28-5%29%2B-sqrt%2825-4%282%29%28-4%29%29%29%2F2%282%29
x+=+%28-%28-5%29%2B-sqrt%2825%2B32%29%29%2F2%282%29
x+=+%28-%28-5%29%2B-sqrt%2867%29%29%2F2%282%29
x+=+%285%2B-sqrt%2867%29%29%2F4
which is short for:
x+=+%285%2Bsqrt%2867%29%29%2F4 or x+=+%285-sqrt%2867%29%29%2F4

So the three solutions are:
x+=+0 or x+=+%285%2Bsqrt%2867%29%29%2F4 or x+=+%285-sqrt%2867%29%29%2F4