Question

find two consecutive positive integers sum of whose squares is 365. (4)

Answer 1

Answer:

Step-by-step explanation:

Let the no. Be X and other be X+1

X² + (X+1)² = 365

X² + X²+1+2x = 365

2x²+ 2x-364=0

2(x²+x-182)= 0

x² + 14x - 13 x -182 = 0

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

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

x = 13

So , x + 1 = 14

Answer 2

Answer :

Two consecutive numbers = 13 and 14

Step-by-step Explanation :-

Let the number be X and other number be X+1

X² + (X+1)² = 365

X² + X²+1+2x = 365

2x²+ 2x-364=0

2(x²+x-182)= 0

x² + 14x - 13 x -182 = 0

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

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

x = 13

Other number x + 1 = 14

Hence, two consecutive numbers are 13 and 14