Question

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

Answer 1

Given:

• Sum of whose squares is 365

To Find:

• Find two consecutive positive integers?

Solution:

Let the two consecutive positive integers be x and x + 1

According to Question:

$\colon\longrightarrow{\sf{ x^2 + (x + 1)^2 = 365}} \\ \\ \\ \colon\longrightarrow{\sf{ x^2 + x^2 + 1 + 2x = 365}} \\ \\ \\ \colon\longrightarrow{\sf{ 2x^2 + 2x - 364 = 0}} \\ \\ \\ \colon\longrightarrow{\sf{ x^2 + x - 182 = 0}} \\ \\ \\ \colon\longrightarrow{\sf{ x^2 + 14x - 13x - 182 = 0}} \\ \\ \\ \colon\longrightarrow{\sf{ x(x + 14) -13(x + 14) = 0}} \\ \\ \\ \colon\longrightarrow{\sf{ (x + 14)(x - 13) = 0}}$

Thus,

x + 14 = 0 or x – 13 = 0,

⇒ x = –14 or x = 13

Since, the integers are Positive.

$\colon{\boxed{\tt\orange{x = 13_{(One \ Integer)} }}}$

$\colon{\boxed{\boxed{\tt\purple{x + 1 = 14_{(Other \ Integer)} }}}}$

Therefore,

• The two consecutive positive integers will be 13 and 14 .

Answer 2

Given :

• Sum of whose squares = 365 .

To find :

• Consecutive positive integers

Solution :

$\sf \: Let ,$

$\sf \: The \: 1st \: Consecutive \: numbers \: be \\ \sf\: x , x + 1$

$\sf Sum \: of \: there \: squares \: will \: be \: \\ \sf \: x² + ( x + 1 )² = 365$

$\;\;\bf{\;\;\blue{x {}^{2} \: + x {}^{2} + 2x \: + 1 = 365}}$

$\;\;\bf{\;\;\blue{2x {}^{2} + 2x \: - 364 = 0 }}$

$\;\;\bf{\;\;\red{x {}^{2} + x - 182 = 0 }}$

$\;\;\bf{\;\;\red{x {}^{2} + 14x - 13x - 182 = 0 }}$

$\;\;\bf{\;\;\green{x {}^{2} + 14x - 13x - 182 = 0}}$

$\;\;\bf{\;\;\green{x(x + 14) - 13(x + 14) = 0}}$

$\;\;\bf{\;\;\blue{(x + 14)(x - 13) = 0}}$

$\;\;\bf{\;\;\red{x = - 14,13}}$

$\;\;\bf{\;\;\green{x = 13}}$

Therefore , the two consecutive positive integer = 13 , 14