Posted by Rachale on Tuesday, August 4, 2009 at 8:58am.
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!
(x-2)/(x-1) + (2)/(x^2-1) = 0
x-2 / x-1+2/(x-1)(x-2) = 0
To check:
0-2 = -2
0-1 = -1
-1/-2 = 0.5
02 – 1 = -1
-1/2 = -0.5
0.5 + -0.5 = 0
1-2 = -1
1-1 = 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?
0-2 = -2
0-1 = -1
-1/-2 = 0.5
0^2 – 1 = -1
-1/2 = -0.5
0.5 + -0.5 = 0
THEN
1-2 = -1
1-1 = 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
(x-2)/(x-1)+(2)/(x^2-1)= 0
it gives a zero as the denominator for each term, which means that each term becomes infinite. The sum of the two terms (left-hand-side) 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!
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