Question

find the two consecutive positive integers,whose sum of their squares is 365.
pls don't copy from Google I want ur own answer pls answer it​

Answer 1

Answer:

13 and 14.

Step-by-step explanation:

Let the required integers are a and a + 1.

Here,  

    ⇒ Sum of their squares is 365

⇒ ( a + 1 )^2 + a^2 = 365

⇒ a^2 + 1 + 2a + a^2 = 365

⇒ 2a^2 + 2a + 1 - 365 = 0

⇒ 2a^2 + 2a - 364 = 0

⇒ 2( a^2 + a - 182 ) = 0

⇒ a^2 + a - 182 = 0

⇒ a^2 + ( 14 - 13 )a - 182 =  0

⇒ a^2 + 14a - 13a - 182 = 0

⇒ a( a + 14 ) - 13( a + 14 ) = 0

⇒ ( a + 14 )( a - 13 ) = 0

⇒ a = - 14    or    a = 13  

     Hence a = 13, since it's +ve, it can't be - 14.

  Required numbers are 13 and 14.

Answer 2

AnswEr:-

Consecutive positive integers = 13 & 14

$\rule{200}{1}$

Given:-

• Sum of squares of two consecutive positive integers = 365

To find:-

• Consecutive positive integers = ?

Solution:-

Let the two consecutive positive integers be n & (n + 1)

According to question:-

⇒ n² + (n + 1)² = 365

⇒ n² + n² + 2n + 1 = 365

[Identity used: (a + b)² = a² + b² + 2ab ]

⇒ 2n² + 2n + 1 - 365 = 0

⇒ 2(n² + n - 182) = 0

⇒ n² + n - 182 = 0

⇒ n² + 14n - 13n - 182 = 0

⇒ n(n + 14) - 13(n + 14) = 0

⇒ (n - 13)(n + 14) = 0

⇒ n = 13 $\sf \: \: or \: \:$ n = -14

∵ n $\neq$ -14 as n is positive integer.

∴ n = 13

$\rule{200}{1}$

$\underline{\bigstar{\sf{Positive\: integers \: :-}}}$

↠ One integer = n = 13

↠ Second integer = n + 1 = 13 + 1 = 14

$\therefore\underline{\textsf{Consecutive Positive Integers ={\textbf{ 13 \& 14 }}}}$