What are the four integral values of x for which #x/(x-2)# has an integral value?

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#x/(x-2)#

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Answer 1 · 2016-03-03T23:06:48

The integer values of #x# are #1,3,0,4#

Lets rewrite this as follows

#x/(x-2)=[(x-2)+2]/(x-2)=1+2/(x-2)#

In order for #2/(x-2)# to be integer #x-2# must be one of the divisors of 2 which are #+-1# and #+-2#

Hence #x-2=-1=>x=1#
#x-2=1=>x=3#
#x-2=-2=>x=0#
#x-2=2=>x=4#

Hence the integer values of x are #1,3,0,4#