Question

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

Answer 1

Answer:

please mark me as brainliest.

Answer 2

Answer:

13 and 14

Step-by-step explanation:

We know that positive integers are the integers that lie on the right hand side of 0 on the number line. Also, we know consecutive numbers mean numbers that are exactly together i.e. right next to each other.

So, consecutive positive integers can be 1,2 ; 2,3 ; 3,4 ; …..

So if we assume the first positive integer as x

Then consecutive positive integer will be x+1

Now we know the sum of squares of two consecutive positive integers is 365

⇒x2+(x+1)2=365

Use the identity (a+b)2=a2+b2+2ab

to open the value in left hand side of the equation

⇒x2+x2+1+2x=365

Shift all the constant values to one side

⇒2x2+2x=365−1

⇒2x2+2x=364

Cancel 2 from both sides of the equation

⇒x2+x=182

⇒x2+x−182=0

Now we use factorization method to solve for value of x

Break the coefficient of ‘x’ in such a way that their sum is equal to coefficient of ‘x’ and their product is equal to the product of coefficient of x2

and the constant term. We can write 182=14×13

We can write 1=14−13

i.e. the coefficient of x

Substitute the value of 1=14−13

in equation (1)

⇒x2+x−182=x2+(14−13)x−182

Open the bracket in RHS of the equation

⇒x2+x−182=x2+14x−13x−182

Take common terms

⇒x2+x−182=x(x+14)−13(x+14)

Combine the factors

⇒x2+x−182=(x+14)(x−13)

Equate factors to 0

x+14=0

and x−13=0

Shift constant values to right hand side

x=−14

and x=13

Since we have both integers positive, we ignore the negative value

⇒x=13

Now we calculate the consecutive positive integer i.e. 13+1=14

The two positive integers having sum of their squares 365 are 13 and 14.