Find two consecutive odd integers such that the square of the first, added to 3 times the second, is 24.

Answer 1

a = first integer

a = 2 k + 1

b = second integer

Becouse odd integers consecutive are consecutive:

b = a + 2 = 2 k + 1 + 2 = 2 k + 3

k is some integer

The square of the first, added to 3 times the second is 24 mean:

a² + 3 b = 24

( 2 k + 1 )² + 3 ∙ ( 2 k + 3 ) = 24

( 2 k )² + 2 ∙ 2 k ∙ 1 + 1² + 3 ∙ 2 k + 3 ∙ 3 = 24

4 k² + 4 k + 1 + 6 k + 9 = 24

4 k² + 10 k + 10 = 24

4 k² + 10 k + 10 - 24 = 0

4 k² + 10 k - 14 = 0

The solutions are :

k = - 7 / 2 and k = 1

k is some integer so k = 1

Yours two consecutive odd integers :

a = 2 k + 1 = 2 ∙ 1 + 1 = 2 + 1 = 3

b = 2 k + 3 = 2 ∙ 1 + 3 = 2 + 1 = 5

Proof:

3² + 3 ∙ 5 = 9 + 15 = 24