Question

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

Answer 1

Answer:

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

Answer:

x^2 + (x+1 )^2 = 365

then what is x

x2 + x2 + 2x + 1 = 365                                                      (a+b)^2

2x2 + 2x - 364 = 0

take 2 common

x2 + 2x - 182

x2 + 14x - 13x - 182

x common

x(x+14)   -13(x+14)

(x+14) (x-13)

x= 13

x + 1 = 14