The rational function #f(x)=1/(x-2)-x/(x^2+3)# has a unique zero. What is the zero of #f(x)#?

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Answer 1 · 2017-07-04T15:03:33

#x=-3/2#

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

The zero of #f# has #f(x)=0#

#therefore1/(x-2)-x/(x^2+3)=0#

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

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

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

#3=-2x#

#x=-3/2#

Answer 2 · 2017-07-04T15:05:19

#-3/2.#

The Zeroes of a function # f ,# is a Set, denoted by #Z_f,# and is defined,

by, #Z_f={x in D_f : f(x)=0, where, D_f"=the Domain of "f.#

We have, #D_f=RR-{3}.#

Now, #f(x)=0 rArr 1/(x-2)-x/(x^2+3)=0.#

# rArr 1/(x-2)=x/(x^2+3).#

# rArr x^2+3=x(x-2)=x^2-2x.#

# rArr -2x=3.#

# rArr x=-3/2.#

#:. Z_f={-3/2}.#