SOLUTION: can you please find the value of 'x' in this equation? (2x-1)^2-x(10x+1) = x(1-x)(1+x)-(2-x)^3 with complete computation please.. thank you so much.

Algebra ->  Equations -> SOLUTION: can you please find the value of 'x' in this equation? (2x-1)^2-x(10x+1) = x(1-x)(1+x)-(2-x)^3 with complete computation please.. thank you so much.       Log On


   

Question 708134: can you please find the value of 'x' in this equation?
(2x-1)^2-x(10x+1) = x(1-x)(1+x)-(2-x)^3
with complete computation please.. thank you so much.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
%282x-1%29%5E2-x%2810x%2B1%29+=+x%281-x%29%281%2Bx%29-%282-x%29%5E3
Alot of FOILing around here
4x%5E2-4x+%2B+1+-+10x%5E2+-+x+=+x%281-x%5E2%29-%284-4x%2Bx%5E2%29%282-x%29
Distribute; get rid of the brackets, change the signs where necessary
4x%5E2+-+10x%5E2+-+4x+-+x+%2B+1+=+%28x+-+x%5E3%29-%288-12x%2B6x%5E2-x%5E3%29
-6x%5E2-5x+%2B+1+=+x+-+x%5E3+-+8+%2B+12x+-+6x%5E2+%2B+x%5E3
Combine like terms
-6x%5E2-5x+%2B+1+=+-x%5E3+%2B+x%5E3+-+6x%5E2+%2B++x+%2B+12x+-+8
-6x%5E2-5x+%2B+1+=+-+6x%5E2+%2B+13x+-+8
variables on the left, numbers on the right
-6x%5E2+%2B+6x%5E2+-+5x+-+13x+=+-8+-+1
-18x = -9
x = -9/-18
x = +.5
:
:
You should confirm this by replacing x with .5 in the original equation.
See that equality reigns supreme.