Find two consecutive natural numbers, sum of whose squares is 365.
Let the two consecutive numbers be x and (x + 1)
Sum of squares of these numbers is 365
⇒ x2 + (x + 1)2 = 365
⇒ x2 + x2 + 2x + 1 = 365
⇒ 2x2 + 2x + 1 – 365 = 0
⇒ 2x2 + 2x – 364 = 0
Take 2 as common factor
⇒ 2 × (x2 + x – 182) = 0
⇒ x2 + x – 182 = 0
⇒ x2 + 14x – 13x – 182 = 0
taking x common from first two terms and – 13 common from next two
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x – 13)(x + 14) = 0
⇒ (x – 13) = 0 or (x + 14) = 0
Therefore x = 13 and x = – 14
if we take x = 13 then the two consecutive numbers whose sum of squares is 365 will be 13 and 14
if we take x = – 14 then the two consecutive numbers whose sum of squares is 365 will be – 13 and – 14