SOLUTION: The cube root of x-1 = the cube root of x^2+ 2x -21

Algebra ->  Square-cubic-other-roots -> SOLUTION: The cube root of x-1 = the cube root of x^2+ 2x -21      Log On


   

Question 27424: The cube root of x-1 = the cube root of x^2+ 2x -21
Answer by mbarugel(146) About Me  (Show Source):
You can put this solution on YOUR website!
We have that the cube root of x-1 = the cube root of x^2+ 2x -21
But since the sube root of a number is unique (as opposed to the square root, which has two possible solutions) this implies that:
x-1 = x^2+ 2x -21
Writing it as a standard quadratic equation, we get:
x%5E2+%2B+2x+-+21+-+x+%2B+1+=+0
x%5E2+%2B+x+-+20+=+0
And now we solve it using the standard procedure:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-20+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-20=81.

Discriminant d=81 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+81+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+81+%29%29%2F2%5C1+=+4
x%5B2%5D+=+%28-%281%29-sqrt%28+81+%29%29%2F2%5C1+=+-5

Quadratic expression 1x%5E2%2B1x%2B-20 can be factored:
1x%5E2%2B1x%2B-20+=+%28x-4%29%2A%28x--5%29
Again, the answer is: 4, -5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-20+%29