Question

find the sum of two consecutive integers whos square is 365​

Answer 1

Let two consecutive positive integer be and .

According to question

Let us split middle term

[ - ve root being rejected ]

Hence required numbers are 13 and 14

Answer 2

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.

∴ x + 1 = 13 + 1 = 14

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

:-) Hope it helps....

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