Competition math problem?

let the numbers be a and b

a^2 + b^2 = 20 => (a+b)^2 - 2ab = 20 ->equation 1

1/a + 1/b = 2 => (a+b)/ab = 2 => a+b = 2ab

substituting this in equation 1,

4(ab)^2 - 2ab -20 =0

=> 2(ab)^2 - ab - 10 = 0

=> 2(ab)^2 - 5ab + 4ab - 10 =0

=> ab(2ab - 5) + 2(2ab -5) =0

=> (ab+2) (2ab-5) = 0

=> ab = -2 which is not possible since both numbers are positive,

or ab = 5/2 which is possible...

hence answer: product is 5/2

let the numbers be x and y

x² + y² = 20 --------- 1

1/x + 1/y = 2 --------- 2

now rearrange eq. 2

( x + y ) / (xy) = 2

x + y = 2xy

now square the eq

x² + y² + 2xy = 4x²y²

20 + 2xy = 4x²y²

let the product xy be 'm'

2m² = 10 + m

2m² - m - 10 = 0

m= -2 , 5/2

the product obvs cant be -2 because x and y are positive

therefore, their product is 5/2