Question 1156496

let's a number be {{{x}}} 

then and its reciprocal is {{{1/x}}} 

if the sum of a number and its reciprocal is {{{37/6}}}, we have

{{{x+1/x=37/6}}} 

{{{(x^2+1)/x=37/6}}} .....cross multiply

{{{6(x^2+1)=37x}}}

{{{6x^2+6=37x}}}

{{{6x^2-37x+6=0}}}.......use quadratic formula


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-(-37) +- sqrt((-37)^2-4*6*6 ))/(2*6) }}}

{{{x = (37 +- sqrt(1369-144 ))/12 }}}

{{{x = (37 +- sqrt(1225 ))/12 }}}

{{{x = (37 +- 35)/12 }}}

solutions:

{{{x = (37 + 35)/12 }}}=>{{{x = 72/12 }}}=>{{{x = 6}}}

or

{{{x = (37 - 35)/12 }}}=>{{{x = 2/12 }}}=>{{{x = 1/6}}}


so, your number is {{{ 6}}} and its reciprocal is {{{ 1/6}}}

or your number is {{{ 1/6}}} and its reciprocal is {{{ 6}}}