How do you solve #1 + 8/(x - 5) = 3/x# and find any extraneous solutions?

1 Answer
Aug 31, 2016

Answer:

There are no Real solutions, but there are Complex solutions:

#x = +-sqrt(15)i#

Explanation:

Given:

#1+8/(x-5) = 3/x#

Note that neither #0# nor #5# can be a solution since they result in division by #0#. So if our derived equations have solutions #0# or #5# then they are extraneous.

Multiply both sides by #x# (possibly introducing an extraneous solution #0#) to get:

#x + (8x)/(x-5) = 3#

Multiply both sides by #(x-5)# (possibly introducing an extraneous solution #5#) to get:

#x(x-5) + 8x = 3(x-5)#

Expand both sides to get:

#x^2-5x+8x = 3x-15#

which simplifies to:

#x^2+3x = 3x-15#

Subtract #3x# from both sides to get:

#x^2 = -15#

This has no Real solutions since #x^2 >= 0# for any Real value of #x#.

If we are interested in Complex solutions, then add #15# to both sides and transpose to get:

#0 = x^2+15 = x^2 - (sqrt(15)i)^2 = (x-sqrt(15)i)(x+sqrt(15)i)#

where #i# is the imaginary unit, with the property that #i^2 = -1#.

Hence solutions #x = +-sqrt(15)i#