Question

find two consecutive odd natural numbers, the sum of whose squares is 394

Answer 1

let us take consecutive =x, x+1
according to question =x^2+(x+1)^2=394
=x^2+(x^2+1^2+2*x*1) (using identity)=394
=x^2+x^2+2+2x=394
=2x^2+2+2x=349
=2x^3=349-2
=2x^3=347
=x^3=347/2
=x=3√347/2
first consecutive =3√347/2
second =3√347/2+1..
thnks:)

Answer 2

Answer:

Let the numbers be x and (x + 2)

• According to the Question :

⇒ x² + (x + 2)² = 394

⇒ x² + x² + 4x + 4 = 394

⇒ 2x² + 4x - 390 = 0

⇒ x² + 2x - 195 = 0

⇒ x² + 15x - 13x - 195 = 0

⇒ (x + 15)(x - 13) = 0

⇒ x = 13

• Required Numbers :

◗ x = 13

◗ (x + 2) = 13 + 2 = 15