Question 479631
Let  the number be {{{n}}} and its reciprocal {{{1/n}}}


given:

the sum of a number and its reciprocal is {{{34/15}}}


{{{n+ (1/n) =34/15}}}.....multiply  thru  by  {{{n}}}


{{{n^2 +1 = (34/15)n}}}.....multiply  thru  by  {{{15}}}


{{{15n^2 +15 = 34n}}}


{{{15n^2 +15 -34n=0}}}.....use quadratic formula


{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}.........{{{a=15}}}, {{{b=15}}} and {{{c=-34}}}

{{{n = (-15 +- sqrt( 15^2-4*15*(-34) ))/(2*15) }}}

{{{n = (-15 +- sqrt( 225+2040 ))/30 }}}


{{{n = (-15 +- sqrt( 2265 ))/30 }}}

{{{n = (-15 +- 47.59)/30 }}}


solutions:

{{{n = (-15 + 47.59)/30 }}}

{{{n = (32.59)/30 }}}

{{{n = 1.086 }}}.......its reciprocal is {{{n = 1/1.086=0.92}}}

or

{{{n = (-15 - 47.59)/30 }}}

{{{n = (-62.59)/30 }}}

{{{n = -2.086 }}}.......its reciprocal is {{{n = -1/2.086=-0.48}}}