Question

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

Answer 1

Let two consecutive positive integer be x and x+1.

According to question,

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

=> x^2 + x^2 + 2x + 1 = 365

=> 2x^2 + 2x - 365 - 1 = 0

=> x^2 + x - 182 = 0

Let us split middle term,

x^2 + 14x - 13x - 182

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

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

∴x = - 14 or x = 13

[ - ve root being rejected ]

Hence required numbers are 13 and 14

I hope the answer helps you.

Answer 2

Answer:

13 and 14

Step-by-step explanation:

Let the smaller of the two consecutive positive integers be x. Then, the second integer will be x + 1.

According to the question,

x² + (x + 1)² = 365

⇒ x² x² + 2 * x * 1 + 1² = 365

⇒ x² + x² + 2x + 1 - 365 = 0

⇒ 2x² + 2x - 364 = 0

⇒ 2 (x² + x - 182) = 0

⇒ x² + x - 182 = 0

Now, we got a quadratic equation,

⇒ x² + x - 182 = 0

⇒ x² + 14x - 13x - 182 = 0

⇒ x ( x + 14 ) - 13 ( x + 14 ) = 0

⇒ ( x - 13 ) ( x + 14 ) = 0

⇒ x = 13 or x = - 14

But x is given to he an odd positive integer. Therefore, x ≠ - 14, x = 13.

Thus, the two consecutive odd integers are 13 and 14.