the sum of two integers is 8, and the sum of their squares is 19 more than their product. Find the integers.

Step-by-step explanation:

a + b = 8 ... a = 8 - b

a^2 + b^2 = ab + 19

(8 - b) ^2 + b^2 = b (8 - b) + 19

(8 - b) (8 - b) + b^2 = 8b - b^2 + 19

64 - 8b - 8b + b^2 + b^2 = 8b - b^2 + 19

2b^2 - 16b + 64 = 8b - b^2 + 19

2b^2 + b^2 - 16b - 8b + 64 - 19 = 0

3b^2 - 24b + 45 = 0

3 (b^2 - 8b + 15) = 0

3 (b - 5) (b - 3) = 0

b - 5 = 0 b - 3 = 0

b = 5 b = 3

when b = 5 when b = 3

a + b = 8 a + b = 8

a + 5 = 8 a + 3 = 8

a = 8 - 5 a = 8 - 3

a = 3 a = 5

ur integers are 5 and 3