Question

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

Answer 1

Answer:

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