Algebra
posted by Rachale on .
Can someone please show me the steps to this problem? I get 0,1 as proposed solutions, but I am not sure about the 1. Thanks
Please solve and check all proposed solutions
(x2)/(x1)+(2)/(x^21)= 0

you're correct, x=1 and x=0

So when we check the problem, would this be acceptable? Does all of this check out? Thanks!
(x2)/(x1) + (2)/(x^21) = 0
x2 / x1+2/(x1)(x2) = 0
To check:
02 = 2
01 = 1
1/2 = 0.5
02 – 1 = 1
1/2 = 0.5
0.5 + 0.5 = 0
12 = 1
11 = 0
0/1 = 0
12 – 1 = 0
0/2 = 0
0+0 = 0
x = 0, 1 
12 – 1 = 0 suppose to be 1^2  1 = 0

02 – 1 = 1 suppose to be 0^2  1 = 0

You're right in suspecting the zero of x=1.
If you subsitiute x=1 into the equation, you will get two indefinite values, hence x=1 has to be rejected.
Note the hint given by the question:
"...and check all proposed solutions ".
All you need to do is to substitute each of the proposed solution into the original equation, if they work out (i.e. LHS=RHS), then the solution is acceptable.
See also:
http://www.jiskha.com/display.cgi?id=1249326304 
Will this check work? Is this correct?
02 = 2
01 = 1
1/2 = 0.5
0^2 – 1 = 1
1/2 = 0.5
0.5 + 0.5 = 0
THEN
12 = 1
11 = 0
0/1 = 0
1^2 – 1 = 0
0/2 = 0
0+0 = 0
x = 0, 1 Does this all check out? 
The first check is OK, so 0 is admitted as a solution.
The second check is not OK because at two places, you have inverted the numerator and denominator, namely:
0/1 = 0 should read 1/0 = ∞
and
0/2 = 0 should read 2/0 = ∞
Since the proposed solution x=1 gives rise to infinite values, x=1 is not admissible as a solution. 
So 1 doesn't work in the solution of the equation at all?

One (1) is rejected as a solution because when substituted into the original equation
(x2)/(x1)+(2)/(x^21)= 0
it gives a zero as the denominator for each term, which means that each term becomes infinite. The sum of the two terms (lefthandside) is thus not equal to zero, so 1 is not admitted as a solution to the given equation. 
Thanks for the explanation!

You are very welcome!