Question

find the two consecutive posetive integer whose sum of squars is 365 .​

Answer 1

Answer:

z

Step-by-step explanation:

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Answer 2

$\huge\boxed{\texttt{\fcolorbox{black}{gold}{Answer}}}$

let one number be $\sf\green{x }$

so, second consecutive number is $\sf\green{x + 1}$

According to Question

⇢ $\sf\green{( x )^2 + ( x + 1 )^2 = 365 }$

⇢ $\sf\green{ x^2 + x^2 +1 +2x = 365 }$

⇢$\sf\green{2x^2 +2x - 364 = 0 }$

⇢ $\sf\green{2( x^2 + x - 182 ) = 0 }$

⇢ $\sf\green{ x^2 +14x -13x -182 = 0/2 }$

⇢ $\sf\green{ x( x + 14 ) - 13( x + 14 ) = 0 }$

⇢$\sf\green{ ( x + 14 ) ( x - 13 ) = 0 }$

$\red{Hence}$

if we take $\sf\green{x-13=0}$

⇢ $\sf\green{ x = 13 }$

so, one number x is $\sf\green{ 13 }$

and second is $\sf\green{x + 1 = 14}$