The sum of a number and its reciprocal is equal to 4 times the reciprocal, What is the number?

3 Answers
Sep 11, 2017

There are two possibilities:

# x=-sqrt(3)# or #x=sqrt(3)#

Explanation:

Suppose the number is #x#

Then the "reciprocal" of this number is #1/x#

We are told that:

"The sum of a number and its reciprocal is equal to 4 times the reciprocal"

Therefore:

# x + 1/x = 4 xx 1/x #

# :. x = 4 xx 1/x - 1 xx 1/x#

# :. x = 3 xx 1/x #

# :. x^2 = 3 #

# :. x = +-sqrt(3 ) #

Sep 11, 2017

#n + 1/n = 4 * 1/n#

on the left side, multiply n by #n/n# (same as multiplying by 1) to get a common denominator so you can add the numerators...

#n^2 /n + 1/n = 4/n#

giving:

#n^2 + 1 = 4#

#n^2 = 3#

so...

#n = +-sqrt(3)#

...check your work, when you can:

#sqrt(3) + 1/sqrt(3) = 4 * 1/sqrt(3)#

#sqrt(3)*sqrt(3)/sqrt(3) + 1/sqrt(3) = 4/sqrt(3)#

#3/sqrt(3) + 1/sqrt(3) = 4/sqrt(3)#

#4/sqrt(3) = 4/sqrt(3)#

(the same sequence also applies for the #-sqrt(3)# case

Sep 11, 2017

I got two possible numbers: #+-sqrt(3)#

Explanation:

Let us call the unknown number #x#; we get:
#x+1/x=4*1/x#
rearrange:
#x^2+1=4#
(with #x!=0#)
#x^2=3#
#x=+-sqrt(3)#