Question

Find two consecutive odd positive integers, sum of whose squares is 290.

Answer 1

Let one of the odd positive integer be x 

Then the other odd positive integer is x + 2 

Their sum of squares = x² + (x + 2)² 

= x² + x² + 4x + 4 

= 2x² + 4x + 4 

Given that their sum of squares = 290 

⇒ 2x² + 4x + 4 = 290 

⇒ 2x² + 4x = 290 - 4 = 286 

⇒ 2x² + 4x - 286 = 0 

⇒ 2(x² + 2x - 143) = 0 

⇒ x² + 2x - 143 = 0 

⇒ x² + 13x - 11x - 143 = 0 

⇒ x(x + 13) - 11(x + 13) = 0 

⇒ (x - 11)(x + 13) = 0 

⇒ (x - 11) = 0, (x + 13) = 0 

Therefore, x = 11 or -13 

According to question, x is a positive odd integer. 

Hence, We take positive value of x 

So, x = 11 and (x + 2) = 11 + 2 = 13 

Therefore, the odd positive integers are 11 and 13.