Question

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

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Answer 1

Answer:

13 and 14

Step-by-step explanation:

As they are consecutive nos.

So let the 1st no be (x)

Then the other no. will be (x+1)

Given that (x)² + (x+1)² = 365

=. x² + x² + 2x + 1 = 365

=. 2x² + 2x + 1 = 365

ON HIT AND TRIAL WE GET

x = 13

Therefore the nos will be (x) i.e. 13 and (x+1) i.e. 14.

Please do mark as BRAINLIEST

Answer 2

Answer:

There is a difference of 1 in consecutive positive integer

Let First interger = x

So, second integer = x+1

Also given that

Sum of squares = 365

(First number)2 + ( Second number)2 = 365

x2 + ( x +1 )2 = 365

x2 + x2 +1 + 2×x×1 = 365

2x2 + 1 + 2x = 365

2x2 + 2x + 1 - 365= 0

2x2 + 2x - 364 = 0

2[ x2 + x - 182 ] = 0

x2 + x - 182 = 0/2

x2 + x- 182 = 0

We factorise by splitting the middle term method

x2 + 14x - 13x - 182 = 0

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

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

x-13=0 , x+14=0

x= 13 , x= -14

So , first number = x= 13

second number x+ 1 = 13 + 1 = 14