Q14 of 83 Page 90

The sum of the squares of two consecutive odd positive integers is 290. Find the numbers.

Let first odd positive integer be x.

⇒ second odd positive integer = x + 2


⇒ x2 + (x + 2)2 = 290


⇒ x2 + x2 + 4 + 4x = 290


⇒ 2x2 + 4x – 286 = 0


Dividing the whole equation by 2, we get –


⇒ x2 + 2x – 143 = 0


⇒ x2 + (13 – 11)x – 143 = 0


⇒ x2 + 13x – 11x – 143 = 0


⇒ x(x + 13) – 11(x + 13) = 0


⇒ (x – 11)(x + 13) = 0


⇒ x = 11 or x = – 13


But x is a positive integer


⇒ x = 11 is the first odd positive integer.


And, second consecutive odd positive integer = x + 2


= 13


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