Question

Find two consecutive positive integers, sum of whose squares is 365.​

Answer 1

Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given statement,

x2 + (x + 1)2 = 365

⇒ x2 + x2 + 1 + 2x = 365

⇒ 2x2 + 2x – 364 = 0

⇒ x2 + x – 182 = 0

⇒ x2 + 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

So, x + 1 = 13 + 1 = 14

Therefore, the two consecutive positive integers will be 13 and 14.

Answer 2

Step-by-step explanation:

Let the numbers be x and x+1

Square of consecutive number: x² and (x+1)²

Sum of square: x² + (x+1)² = 365

Lets solve

x²+x²+2x+1 = 365

2x²+2x-364=0

x²+x-182=0

(Dividing by 2)

Factorise,

x²+14x-13x-182=0

x(x+14)-13(x+14)=0

(x+14)(x-13)=0

∴x=-14 and+13

Since numbers are positive integers,x=13

So, required consecutive numbers are 13 and 14.