Question 31243
Hello!
Given a number X, we call 1/X its reciprocal. So, for example 1/2 is the reciprocal of 2.

So we have that the sum of a number and its reciprocal (X + 1/X) is 10/3:

{{{X + 1/X = 10/3}}}

Multiply everything by X in order to get rid of the 1/X:
{{{X*X + 1/X*X = 10/3*X}}}
{{{X^2 + 1 = (10/3)X}}}
{{{X^2 - (10/3)X + 1 = 0}}}

This is a quadratic equation, which can be solved using the quadratic formula:

{{{x[12] = (-b+-sqrt( b^2-4ac ))/2a }}}

So we have:
{{{sqrt((-10/3)^2 - 4*1*1) = sqrt(100/9-4) = sqrt(64/9) = 8/3}}}


{{{x[12] = ( -(-10/3) +- (8/3) )/(2*1) = ( (10/3) +- (8/3) )/2}}}

So the two possible numbers that solve your problem are:

{{{x[1] = 3}}}
{{{x[2] = 1/3}}}


I hope this helps!
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