Question

Find two consecutive positive integer whossum is 365​

Answer 1

Let two consecutive positive integer be x and x+1.

According to question

x2 + (x+1)2=365

=x2+x2+2x+1=365

=2x 2 +2x-364=0

=x2+x-182=0

Let us split middle term

x2+14x-13x-182=0

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

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

x=14 or x= - 13

[ - ve root being rejected ]

Hence required numbers are 13 and 14

Answer 2

Answer:

INTERGERS ARE 13 AND 14

Step-by-step explanation:

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