What is the solution set for #8/(x+2)=(x+4)/(x-6)#?

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Answer 1 · 2015-08-12T19:10:14

There are no real solutions and two complex solutions #x=1\pm i sqrt(55)#

First, cross multiply to get #8(x-6)=(x+2)(x+4)#. Next, expand to get #8x-48=x^2+6x+8#. Now rearrange to obtain #x^2-2x+56=0#.

The quadratic formula now gives solutions

#x=(2\pm sqrt(4-224))/2=1\pm 1/2 sqrt(-220)#

#=1\pm 1/2 i sqrt(4)sqrt(55)=1\pm isqrt(55)#

These are definitely worth checking in the original equation. I'll check the first and you can check the second.

The left-hand-side of the original equation, upon substitution of #x=1+i sqrt(55)# becomes:

#8/(3+isqrt(55))=(8(3-isqrt(55)))/(9+55)=3/8-i sqrt(55)/8#

Now do the same substitution on the right-hand-side of the original equation:

#(5+isqrt(55))/(-5+isqrt(55))=((5+isqrt(55)) * (-5-isqrt(55)))/(25+55)#

#=(-25-10isqrt(55)+55)/80=3/8-i sqrt(55)/8#

It works! :-)