SOLUTION: Hi, can you please help me?? 1. I need to solve the following equation: 8x^3+1=0 ---> I tried doing the sum of two cubes, and ended up with x= -1/2, 1/2, and 1/2...can you p

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi, can you please help me?? 1. I need to solve the following equation: 8x^3+1=0 ---> I tried doing the sum of two cubes, and ended up with x= -1/2, 1/2, and 1/2...can you p      Log On


   

Question 404639: Hi, can you please help me??
1. I need to solve the following equation:
8x^3+1=0 ---> I tried doing the sum of two cubes, and ended up with x= -1/2, 1/2, and 1/2...can you please tell me if I'm doing it wrong, and show me how to do it correctly?
Thank you so much!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
8x%5E3%2B1+=+0
The sum of cubes is a way to solve this problem. But if used correctly you will not get all the answers you found. Since you did not post your work I cannot see where you went wrong.

Rewriting your equation as a sum of cubes we get:
%282x%29%5E3+%2B+1%5E3+=+0
Now we can factor according to the sum of cubes pattern: a%5E3%2Bb%5E3+=+%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29 with "a" being 2x and "b" being 1:
%28%282x%29%2B%281%29%29%28%282x%29%5E2-%282x%29%281%29%2B%281%29%5E2%29+=+0
which simplifies to:
%282x%2B1%29%284x%5E2-2x%2B1%29+=+0
The second factor does not factor easily. From the Zero Product Property we know that this product can be zero only if one of the factors is zero. So:
2x%2B1+=+0 or 4x%5E2-2x%2B1+=+0
The first equation is simple to solve, We get
x = -1/2
The second one will require the Quadratic Formula:
x+=+%28-%28-2%29+%2B-+sqrt%28%28-2%29%5E2-4%284%29%281%29%29%29%2F2%284%29
which simplifies as follows:
x+=+%28-%28-2%29+%2B-+sqrt%284-4%284%29%281%29%29%29%2F2%284%29
x+=+%28-%28-2%29+%2B-+sqrt%284-16%29%29%2F2%284%29
x+=+%28-%28-2%29+%2B-+sqrt%28-12%29%29%2F2%284%29
At this point we find that there is a negative number inside the square root. This means that the two solutions we will get from this equation will be complex numbers. If you do not know about complex numbers or if you know you are supposed to find only real solutions, then we are done. There is just one real solution: x = -1/2. But if you know about complex numbers then we can continue:
x+=+%282+%2B-+sqrt%28-12%29%29%2F8
x+=+%282+%2B-+sqrt%28-1%2A4%2A3%29%29%2F8
x+=+%282+%2B-+sqrt%28-1%29%2Asqrt%284%29%2Asqrt%283%29%29%2F8
sqrt%28-1%29+=+i
x+=+%282+%2B-+i%2A2%2Asqrt%283%29%29%2F8
x+=+%282%281%2B-+i%2Asqrt%283%29%29%29%2F%284%2A2%29
x+=+%28cross%282%29%281%2B-+i%2Asqrt%283%29%29%29%2F%284%2Across%282%29%29%29
x+=+%281%2B-+i%2Asqrt%283%29%29%2F4
In long form this is:
x+=+%281%2B+i%2Asqrt%283%29%29%2F4 or x+=+%281-+i%2Asqrt%283%29%29%2F4
Writing these in proper form for complex numbers:
x+=+1%2F4%2B+%28i%2Asqrt%283%29%29%2F4 or x+=+1%2F4-+%28i%2Asqrt%283%29%29%2F4
x+=+1%2F4%2B+%28sqrt%283%29%2F4%29i or x+=+1%2F4+%2B+%28-sqrt%283%29%2F4%29i
So if complex solutions are to ne included, there are three solutions:
x+=+-1%2F2 or x+=+1%2F4%2B+%28sqrt%283%29%2F4%29i or x+=+1%2F4+%2B+%28-sqrt%283%29%2F4%29i

P.S. If only real solutions were desired, then there was a much quicker way:
8x%5E3%2B1+=+0
8x%5E3+=+-1
root%283%2C+8x%5E3%29+=+root%283%2C+-1%29
2x = -1
x = -1/2

P.P.S. In response to the question in your thank you note...
It was a good idea to try to factor 4x%5E2-2x%2B1. But, as I mentioned above, 4x%5E2-2x%2B1 will not factor. If you factored it you must have made a mistake.

I consider the Quadratic Formula a last resort. The plus/minus, the square root and the complexity of the formula all make it difficult to work with. So I use it only if factoring does not work or if factoring seems to be even more work than the formula. I used the formula in this problem because 4x%5E2-2x%2B1 will not factor.