Question

Find two consecutive odd integers such that the square of the first,

added to 3 times the second, is 34.

Answer 1

-7 and -5

Explanation:

let a be an odd integer.

Then ‘a’ can be written as :

a = 2p + 1 ; where p is an integer.

The next odd number to ‘a’ is :

a + 2 = 2p + 1 = 2p + 3

The statement “the square of the first, added to 3 times the second,

is 34”

means

a² + 3(a + 2) = 34

………………………………………………

Solving the equation :

a² + 3(a + 2) = 34

⇔ (2p + 1)² + 3(2p + 3) = 34

⇔ 4p² + 4p + 1 + 6p + 9 = 34

⇔ 4p² + 10p + 10 = 34

⇔ 4p² + 10p - 24 = 0

⇔ 2p² + 5p - 12 = 0

⇔ 2p² + 5p +3p - 3p - 12 = 0

⇔ 2p² + 8p - 3(p + 4) = 0

⇔ 2p(p + 4) - 3(p + 4) = 0

⇔ (p + 4) (2p - 3) = 0

⇔ p + 4 = 0 or 2p - 3 = 0

⇔ p = -4 or p = 3/2

(3/2 to be rejected because not an integer)

Then

p = -4

Calculating a and a+2 :

a = 2p + 1 = 2(-4) + 1 = -7

a + 2 = 2p + 3 = 2(-4) + 3 = -5

Verification :

(-7)² + 3(-5) = 49 - 15 = 34