Question

find two consecutive positive integers sum of whose squares is 365.I will mark them as brain list if the answer is correct
hope you help

Answer 1

 
Find two consecutive positive integers, sum of whose squares is 365.
 
Solution:
Let first number be x
Let second number be (x+1)
 
According to given condition, we have
x2+(x+1)2=365                                                    {(a+b)2=a2+b2+2ab}
 
⇒x2+x2+1+2x=365

⇒2x2+2x−364=0
 
Dividing equation by 2, we get
x2+x−182=0
 
⇒x2+14x−13x−182=0
 
⇒x(x+14)−13(x+14)=0
 
⇒(x+14)(x−13)=0
 
⇒x=13,−14
 
Therefore first number = 13   {We discard -14 because it is given that number is positive).
Second number = x+1=13+1=14
Therefore two consecutive positive integers are 13 and 14 whose sum of squares is equal to 365.



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Answer 2

let one integer =x
another =x+1


ACCORDING TO THE QUESTION,


x+x+1=365
2x+1=365
2x=364
$x = \frac{364}{2}$
$x = 182$
x+1=183


PLEASE MARK IT AS BRAINLIST