Question

product of two number is 56 when sum is added to sum of there squar we get 128 find the two numbers

Answer 1

xy=56

x+y+x2+y2=128

(x+y)+(x+y)^2 - 2xy=128

lets assume x+y=k

k+k^2-2(56)=128, from xy=56

k^2 +k-240=0

k=15,-16

so k=15

x+y=15

xy=56

x=8,y=7

Answer 2

xy = 56

x = 56/y

x + y + x² + y² = 128

x+ y + x² + y² + 2xy - 2xy = 128

(x + y) + (x+y)² - 2xy = 128

(x + y) + (x+y)² - 2x 56 = 128

(x+y) + (x +y)² = 128 + 112

(x+y)² + (x+y) = 240

if x+ y = a

a² + a = 240

a² + a - 240 = 0

a² + 16a - 15a - 240 = 0

a (a + 16) - 15 (a + 240) = 0

(a - 15) (a+16) = 0

a = 15 or -16

since we will get irrational roots, we can't be minus number so we will take 15

a = x + y = 15

x + 56/x = 15

x² - 15x + 56

x² - 8x - 7x + 56

x(x -8) - 7(x - 8)

(x -7) (x  - 8)

x = 7 , 8

so numbers are 8 and 7