The sum of squares of two numbers is 80 and the square of
difference between the two numbers
is 36. Find the product of two
numbers .

the product of the 2 numbers is 22

Step-by-step explanation:

x² + y² = 80

(x - y)² = 36

=>

x - y = 6

or y - x = 6

let's start with the first one x-y=6

x = 6 + y

=>

(6+y)² + y² = 80

y² + 6y +6y + 36 + y² = 80

2y² + 12y + 36 = 80

2y² + 12y - 44 = 0

y² + 6y - 22 = 0

the solution of a quadratic equation

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case here we have an squadron in y.

a=1

b=6

c=-22

(-6 ± sqrt(36 + 4×22))/2 = (-6 ± sqrt(36+88))/2 =

= (-6 ± sqrt(124))/2 = (-6 ± sqrt(4×31))/2 =

= (-6 ± 2×sqrt(31))/2 = -3 ± sqrt(31)

y1 = -3 + sqrt(31)

y2 = -3 - sqrt(31)

=>

x1 = 6 + -3 + sqrt(31) = 3 + sqrt(31)

x2 = 3 - sqrt(31)

control :

(-3 + sqrt(31))² + (3 + sqrt(31))² = 80

9 - 3 sqrt(31) - 3 sqrt(31) + 31 + 9 +3 sqrt(31) + 3 sqrt(31) + 31 = 80

9 + 31 + 9 +31 = 80

18 + 62 = 80

80 = 80 correct

solving now for y-x=6

delivers exactly the same calculations, just with x and y trading places.

so, the resulting 2 number pairs are the same.

the product of the 2 numbers :

(3 + sqrt(31))(-3 + sqrt(31)) = -9 - 3 sqrt(31) + 3 sqrt(31) + 31 =

= -9 + 31 = 22

(3 - sqrt(31))(-3 - sqrt(31)) = -9 + 3 sqrt(31) - 3 sqrt(31) + 31 =

= 22

so, the product is the same in both cases.