find two consecutive positive integers, sum of whose squares is 365 .
Share
find two consecutive positive integers, sum of whose squares is 365 .
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Given:
[tex] \\ [/tex]
To Find:
[tex] \\ [/tex]
Solution:
Let the two consecutive positive integers be x and x + 1
According to Question:
[tex] \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}} \\ [/tex]
Thus,
x + 14 = 0 or x – 13 = 0,
⇒ x = –14 or x = 13
Since, the integers are Positive.
[tex] \colon{\boxed{\tt\orange{x = 13_{(One \ Integer)} }}} [/tex]
[tex] \colon{\boxed{\boxed{\tt\purple{x + 1 = 14_{(Other \ Integer)} }}}} [/tex]
Therefore,
Given :
To find :
Solution :
[tex]\sf \: Let , [/tex]
[tex]\sf \: The \: 1st \: Consecutive \: numbers \: be \\ \sf\: x , x + 1 [/tex]
[tex]\sf Sum \: of \: there \: squares \: will \: be \: \\ \sf \: x² + ( x + 1 )² = 365 [/tex]
[tex]\;\;\bf{\;\;\blue{x {}^{2} \: + x {}^{2} + 2x \: + 1 = 365}} [/tex]
[tex]\;\;\bf{\;\;\blue{2x {}^{2} + 2x \: - 364 = 0 }} [/tex]
[tex]\;\;\bf{\;\;\red{x {}^{2} + x - 182 = 0 }} [/tex]
[tex]\;\;\bf{\;\;\red{x {}^{2} + 14x - 13x - 182 = 0 }} [/tex]
[tex]\;\;\bf{\;\;\green{x {}^{2} + 14x - 13x - 182 = 0}} [/tex]
[tex]\;\;\bf{\;\;\green{x(x + 14) - 13(x + 14) = 0}} [/tex]
[tex]\;\;\bf{\;\;\blue{(x + 14)(x - 13) = 0}} [/tex]
[tex]\;\;\bf{\;\;\red{x = - 14,13}} [/tex]
[tex]\;\;\bf{\;\;\green{x = 13}} [/tex]
Therefore , the two consecutive positive integer = 13 , 14