Question

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

Answer 1

Let integers be x and (x+1)

So,
x[tex] x^{2} + (x+1)^{2} = 365
or, x^{2} + x^{2} + 2x + 1 = 365
or, 2 x^{2} + 2x - 364 = 0 [/tex]
or, [tex] x^{2} + 14x - 13x - 182 = 0
or, (x +14)(x-13) = 0[/tex]
or, x = 13

So, 2 positive integers are 13 and (13+1) = 14

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.....:)