How do you solve #1-13/x+36/x^2=0# and check for extraneous solutions?

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Answer 1 · 2017-05-15T13:19:55

#x=9# or #x=4# and there are no extraneous solutions.

Observe that as in denominator, we just have #x# or #x^2# only restrictions on domain is that #x!=0#

Let #1/x=u# then #1-13/x+36/x^2=0# can be written as

#36u^2-13u+1=0#

or #36u^2-9u-4u+1=0#

or #9u(4u-1)-1(4u-1)=0#

or #(9u-1)(4u-1)=0#

i.e. #u=1/9# or #u=1/4#

Hence #x=9# or #x=4# and there are no extraneous solutions.