Question

Find two consecutive positive integers, sum of whose square is 365.

Answer 1

Hey there,

Your answer.

-》Let the consecutive positive integers be xand x + 1.
Given that x2 +(x+1)2 = 365
= x2 + x2 +1+2x = 365
= 2x2 +2x – 364 = 0
= x2 + x – 182 = 0
= x2 + 14x – 13x – 182 = 0
= x(x+14) – 13(x+14) = 0
= (x+14)(x – 13) = 0
Either x + 14 = 0 or x − 13 = 0, i.e., x = −14 or x= 13
Since the integers are positive, x can only be 13.
so, x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.

:-) Hope it helps......

Answer 2

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Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given statement,

$\tt x^2 + (x + 1)^2 = 365$

$\tt ⇒ x^2 + x^2 + 1 + 2x = 365$

$\tt ⇒ 2x^2 + 2x – 364 = 0$

$\tt ⇒ x^2 + x – 182 = 0$

$\tt ⇒ x^2 + 14x – 13x – 182 = 0$

$\tt ⇒ x(x + 14) -13(x + 14) = 0$

$\tt ⇒ (x + 14)(x – 13) = 0$

Thus, either, x + 14 = 0 or x – 13 = 0, [/tex]

$\tt ⇒ x = – 14 or x = 13$

since, the integers are positive, so x can be 13, only.

So, x + 1 = 13 + 1 = 14

Therefore, the two consecutive positive integers will be 13 and 14

Hope it's Helpful.....:)