SOLUTION: The negative reciprocal of the sum of all values of x that satisfy the equation {{{ (x-sqrt(x+1))/(x+sqrt(x+1)) =11/5 }}} is: a) 64/9 b) -9/64 c) 9/8 d) -1/8 e) 8/9

Algebra ->  Equations -> SOLUTION: The negative reciprocal of the sum of all values of x that satisfy the equation {{{ (x-sqrt(x+1))/(x+sqrt(x+1)) =11/5 }}} is: a) 64/9 b) -9/64 c) 9/8 d) -1/8 e) 8/9      Log On


   

Question 1105913: The negative reciprocal of the sum of all values of x that satisfy the equation +%28x-sqrt%28x%2B1%29%29%2F%28x%2Bsqrt%28x%2B1%29%29+=11%2F5+ is:
a) 64/9 b) -9/64 c) 9/8 d) -1/8 e) 8/9
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You could try to solve +%28x-sqrt%28x%2B1%29%29%2F%28x%2Bsqrt%28x%2B1%29%29+=11%2F5+
by transforming it into
11%28x%2Bsqrt%28x%2B1%29%29+=+5%28x-sqrt%28x%2B1%29%29 ,
11x%2B11sqrt%28x%2B1%29+=+5x-5sqrt%28x%2B1%29 ,
11x-5x=-11sqrt%28x%2B1%29-5sqrt%28x%2B1%29 , and
6x=-16sqrt%28x%2B1%29 .
Then you would have to square both sides,
solve the resulting equation,
and eliminate extraneous solutions, if any.
(There will be an extraneous solution,
and the whole process is not fun).

AN EASIER WAY:
Let us define
y=sqrt%28x%2B1%29%3E=0 .
y%5E2=x%2B1
y%5E2-1=x .
With that change of variable, the equation becomes
+%28y%5E2-1-y%29%2F%28y%5E2-1%2By%29+=11%2F5+
Now we can solve for y and take any solution such that y%3E=0 .
+%28y%5E2-1-y%29%2F%28y%5E2-1%2By%29+=11%2F5+
11%28y%5E2-1%2By%29=5%28y%5E2-1-y%29
11y%5E2-11%2B11y=5y%5E2-5-5y
11y%5E2-5y%5E2%2B5y%2B11y-11%2B5=0
6y%5E2%2B16y-6=0 .
We can divide both sides of the equal sign by 2 to simplify it to
3y%5E2%2B8y-3=0 .
We could solve it by factoring as follows.
3y%5E2%2B8y-3=0 --> %283y-1%29%28y%2B3%29=0 --> system%28x=-3%2C%22or%22%2Cx=1%2F3%29
Using the quadratic formula or solving by "completing the square" would yield the same results.
As we need y%3E=0 , our solution is y=1%2F3 .
There is no other solution.
So, x=y%5E2-1=1%2F9-1=-8%2F9 is the only solution, and the sum of all solutions.
The negative reciprocal of -8%2F9 is highlight%289%2F8%29 .