Two numbers have a sum of 8 and a product of 11. Find the sum of their reciprocals?

Its not a trick question, but im just not sure how to do it.Two numbers have a sum of 8 and a product of 11. Find the sum of their reciprocals?
The question creates two equations, namely:


x+y=8 and


xy=11





Substituting x=8-y in for x in the second equation we have:





(8-y)(y)=11


8y-y^2=11


0=y^2-8y+11


since this equation doesn't factor down, one must use the quadratic equation. Solutions for Y are : 4 (plus or minus) Sqrt(5).


We'll assume they both have validity.





Using Y= the positive root, that makes x=8-4-sqrt(5)


Using Y= the Negative root, makes x=8-4+sqrt(5)





and the sum of the reciprocals is either 8/11 either way


Two numbers have a sum of 8 and a product of 11. Find the sum of their reciprocals?
let x be one number and y the other





x+y=8 --%26gt; y = 8 - x


xy = 11


so


x(8 - x) = 11


8x - x^2 - 11 = 0





use the quadratic formula, and you get


x = 4 +- sqrt(5)


so, the 2 numbers are


4 + sqrt(5) and 4 - sqrt(5)





the sum of the reciprocals is


1/(4 + sqrt(5)) + 1/(4 - sqrt(5))





so,





(4 - sqrt(5))/11 + (4+sqrt(5))/11, which is





8/11
1/a + 1/b = (a+b)/ab = 8/11