SOLUTION: I'm supposed to find the solutions of x for the equation: 27x^3=(x+5)^3 I tried using the difference of cubes where a=(x+5)^3 and b=27x^3 but running through it gave a polyno

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I'm supposed to find the solutions of x for the equation: 27x^3=(x+5)^3 I tried using the difference of cubes where a=(x+5)^3 and b=27x^3 but running through it gave a polyno      Log On


   

Question 632223: I'm supposed to find the solutions of x for the equation:
27x^3=(x+5)^3
I tried using the difference of cubes where a=(x+5)^3 and b=27x^3 but running through it gave a polynomial to the 6th power, which doesn't do me much good. Please help.
Found 3 solutions by reviewermath, solver91311, Earlsdon:
Answer by reviewermath(1029) About Me  (Show Source):
You can put this solution on YOUR website!
Using the difference of two cubes where a%5E3+=+%28x%2B5%29%5E3 and b%5E3=27x%5E3.
a%5E3-b%5E3+=+0+
%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29+=+0
%28%28x%2B5%29+-+3x%29%2A%28%28x%2B5%29%5E2+%2B+3x%28x%2B5%29+%2B+%283x%29%5E2%29+=+0
%28-2x+%2B+5%29%2A%28x%5E2+%2B+10x+%2B+25+%2B+3x%5E2+%2B+15x+%2B+9x%5E2%29+=+0
%28-2x+%2B+5%29%2A%2813x%5E2+%2B+25x+%2B+25%29+=+0
+-2x+%2B+5+=+0 or 13x%5E2+%2B+25x+%2B+25+=+0
Solve: -2x+%2B+5+=+0
highlight%28x+=+5%2F2%29
Solve: 13x%5E2+%2B+25x+%2B+25+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 13x%5E2%2B25x%2B25+=+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=%2825%29%5E2-4%2A13%2A25=-675.

The discriminant -675 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -675 is + or - sqrt%28+675%29+=+25.9807621135332.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+13%2Ax%5E2%2B25%2Ax%2B25+%29

There is only one real solution, x = 5/2.
The other two roots are imaginary numbers as computed above.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Your error, I think, was stating that and when in fact you should have started with and , which would lead you to and

Then using the pattern to write:



You can take it from here. You should get a linear binomial factor and a quadratic trinomial on which you will need to use the quadratic formula. The net result will be one real root and a pair of complex conjugates.

John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Try a simpler approach:
27x%5E3+=+%28x%2B5%29%5E3 Rewrite as:
%283x%29%5E3+=+%28x%2B5%29%5E3 Take the cube root of both sides.
3x+=+x%2B5 Subtract x from both sides.
2x+=+5 Finally, divide both sides by 2.
x+=+2.5