How do you solve #x^2 / (x+3) - 5 / (x+3) = 0# and find any extraneous solutions?

1 Answer
Sep 27, 2016

Answer:

#x = +-sqrt5#

Explanation:

As there are only two fractions - let's move one to the other side.
Then we can cross-multiply..

#color(blue)(x^2/(x+3) = 5/(x+3)" "larr (x+3 !=0))#

#x^2(x+3) = 5(x+3)#

#x^3+3x^2= 5x+15" "larr# no fractions! Make equal to 0

#x^3+3x^2 -5x-15 =0" "larr# find factors by grouping

#x^2(x+3) -5(x+3)=0" "larr# common bracket

#(x+3)(x^2-5) =0#

If #x+3 = 0 " "rarr x =-3# (extraneous solution #x-3 !=0) #

If #x^2 -5 = 0 " "rarr x = +-sqrt5 " "larr# the only solutions.

However if we consider the fraction right at the beginning, we could have arrived at the answer in a much quicker way!

#color(blue)(x^2/(x+3) = 5/(x+3)" "larr (x+3 !=0))#

The denominators are equal, so the numerators are equal..

#x^2 = 5#

#x = +-sqrt5#

#x+3 !=0, " so " x !=-3#