How do you solve #2/(x-2) + 7/(x^2-4) = 5/x#?

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Answer 1 · 2016-05-23T15:46:35

#x= 5" and " -4/3#

Known: #a^2-b^2=(a-b)(a+b)#

So #x^2-4->x^2-2^2 -> (x-2)(x+2)#

Write the given equation as:

#2/(x-2)+7/[(x-2)(x+2)]=5/x#

#=>[2(x+2)+7]/[(x-2)(x+2)]=5/x#

#=>[2x+4+7]/[(x-2)(x+2)]=5/x#

Cross multiplying

#x(2x+11)=5(x-2)(x+2)#

#x(2x+11)=5(x^2-4)#

#2x^2+11x=5x^2-20#

#3x^2-11x-20=0#

not all the solutions are whole numbers

Using: #y=ax^2+bx+c# where a=3; b= -11; c=-20

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#=>x=(11+-sqrt(11^2-4(3)(-20)))/(2(3))#

#x=(11+-sqrt(121+240))/6#

#x= (11+-19)/6#

#x= 5" and " -4/3#