Question

\[\left( 1 \over x +1 \right)-2\left( 1\over x+1 \right)-8=0\]

Answer 1

if we multiply it all by (x+1) we can clear out the fractional parts

Answer 2

1-2-8(x+1) = 0

Answer 3

1 -2 -8x -8 = 0

-9 -8x = 0

-8x = 9

x = -9/8 maybe?

Answer 4

:) that does make a difference; and please refrain from typing in profanity. We want to keep the place clean.

Answer 5

substitute [1/(x+1)] with a different variable like; b

then it becomes b^2 -2b -8 = 0, and we can solve for b

Answer 6

(b-4) (b+2) ; b= 4 and -2

since b= 1/(x+1) this means

\[\frac{1}{x+1}=4 \ or -2\]

Answer 7

ah i see.
answer is: -3/2 and -3/4

Answer 8

\[\frac{1}{x+1}=4\iff 1 = 4x +4\]

\[4x = -3 \iff x = -3/4\]
also
\[\frac{1}{x+1}=-2\iff 1= -2x-2\]
\[3 = -2x \iff x = -3/2\]
yes, very good :)

Answer 9

why can't we take \[\left( 1 \over x + 1 \right)^2\] and write it as: \[\left( 1 \over x+1 \right)\left( 1 \over x+1 \right)\]?

Answer 10

we can, but it really doesnt lead to an easier solution to the problem.

Answer 11

it's just the way that i learned and was wondering if i was doing it a way that was wrong... thanks for clearing it up on that :)

Answer 12

youre welcome :)

Answer 13

or... how am i doing it wrong?

Answer 14

since that is neither equal to -3/2 or -3/4 .....

I would have to see how you are trying to solve it.

Answer 15

n^2 -2n -8 = 0 ; are you attempting to factor the first 2?

n(n-2) -8 = 0

n(n-2) = 8 ??

Answer 16

i wrote it as: \[\left( 1 \over x+1 \right)\left( 1 \over x+1 \right)\] then used common denominator. it leaves me with 1*1 which doesn't seem right...

Answer 17

the common denom for all of them would then be what: x^2 +2x +1 right?

Answer 18

that would become the new denominator from 1/(x+1) times 1/(x+1)?

Answer 19

\[\frac{1}{x^2+2x+ 1}-\frac{2(x+1)}{x^2+2x+1}-\frac{8(x^2+2x+1)}{x^2+2x+1}\]

Answer 20

which then amounts to just focusing on the top since that is what makes a fraction = 0

\[1-2x-2-8x^2-16x-8\]

Answer 21

hmmm... i'm trying to understand. you have been really helpful. i am going to work through this problem a time or two more...

Answer 22

good luck with it :)