How do you solve #(6/(x^2-1)) - (1/(x+1)) = 1/2#?

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Answer 1 · 2016-04-06T10:13:50

#x~~ -4.464" or "2.464# to 3 decimal paces

#x^2-1# can be written as #x^2-1^2#

Compare this to #a^2+b^2=(a+b)(a-b)#

We can use this to our advantage

Write the given equation as:

#6/(x^2-1^2) - 1/(x+1)=1/2#

By substitution this is

#6/((x+1)(x-1)) - 1/(x+1)=1/2#

#(6-(x-1))/((x+1)(x-1))=1/2#

Multiply both sides by 2

#(12-2(x-1))/((x+1)(x-1))=1#

#12-2(x-1)=(x+1)(x-1)#

#12-2x+2=x^2-1#

#x^2+2x-11=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now solve as a quadratic -

Completing the square method (skipping many steps)

#(x+1)^2 -12=0#

#x+1= +-sqrt(12)#

#x= -1+-2sqrt(3)#

#x~~ -4.464" or "2.464# to 3 decimal paces