How do you find the zeroes of #F(x) = 1 + 3/(x^2 - 4)#?

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Answer 1 · 2015-07-19T15:56:08

Reformulate and factorize to find: #F(x) = 0# when #x=+-1#

#F(x) = 1+3/(x^2-4)#

#= (x^2-4)/(x^2-4)+3/(x^2-4)#

#= (x^2-4+3)/(x^2-4)#

#=(x^2-1)/(x^2-4)#

#=(x^2-1^2)/(x^2-2^2)#

#=((x-1)(x+1))/((x-2)(x+2))#

using the difference of squares identity:

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

So #F(x) = 0# when #x=+-1#