Question

Find two consecutive integers sum of whose squares is 365​

Answer 1

Explanation:

Let the two consecutive positive integers be x and x+1

Then,

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

⇒x^2 +x^2 +2x+1=365

⇒2x^2 +2x−364=0

⇒x^2+x−182=0

Using the quadratic formula, we get

x={ −1 ± √[1+728] }/2

⇒[ −1 ±27]/2

⇒x=13 and x=−14

But x is given to be a positive integer. ∴x



=−14

Hence, the two consecutive positive integers are 13 and 14.

Answer 2

Explanation:

ANSWER

Let the consecutive numbers be x,x+1

Sum of their squares x

2

+(x+1)

2

=365

x

2

+x

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=−14,13

x=13

two consecutive positive integers=13,14