Question

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

Answer 1

Let the smaller number among the two unknown numbers be x.

Since the numbers are consecutive, the other number is (x+1)

Their squares add up to give 365

So, we can write

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

$2x^2+2x+1 = 365$

$2x^2+2x-364 = 0$

$x^2+x-182 = 0$

$x^2+14x-13x-182 = 0$

$x(x+14)-13(x+14) = 0$

$(x+14)(x-13) = 0$

x = -14 OR x = 13

Since it is specified in the question that the numbers are positive, we can reject the negative value.

So the numbers are 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.....:)