Question

$\huge\underline\mathfrak\green{Question}$


=> Find two consecutive positive integers, sum of whose squares is 365.


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Answer 1

Answer:

let the no. be  x and x+1

 

then we have x² +(x+1)² =365

                     2x² +2x - 364 =0

                     x² + x - 182=0

                     x² +14x-13x-182=0

                     (x+14)(x-13)=0

  That gives x=13 Avoiding negative value one number is 13 and other one 14.

#Hope it helps ✨

Answer 2

$\huge{\underline{\underline{\mathfrak{\purple{Answer}}}}}$

$\huge\boxed{13\:and\:14}$

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$\huge{\underline{\underline{\mathfrak{\purple{Explanation}}}}}$

Let 1st consecutive positive integer be n.

Therefore, 2nd consecutive positive integer be n+1.

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According to question,

(n)²+(n+1)² = 365

⇒ n² + n² + 1 + 2n = 365 [ By using (a+b)² = a²+b²+2ab ]

⇒ 2n² + 2n = 365 - 1

⇒ 2n² + 2n = 364

⇒2n² + 2n - 364 = 0

⇒n² + n - 182 = 0

By splitting middle term :

⇒ n² + 14n - 13n - 182 = 0

⇒ n( n+14) - 13(n+14) = 0

⇒ (n-13) = 0 and (n+14) = 0

⇒n= 13,-14

[ negative root must be rejected as only positive consecutive integers are needed as mentioned in the question ]

Hence, n = 13

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Therefore, 1st consecutive positive integer = n = 13

2nd consecutive positive integer = n+1 = 13+1 = 14