Question

The sum of squares of two consecutive odd integers is 394 find the numbers

Answer 1

Step-by-step explanation:

Let the two consecutive integers be x and (x+2)

By problem,

x^2+(x+2)^2= 394

or,x^2 +x^2+4x+4=394

or, 2x^2+2x-390=0

or, x^2+x-195=0

or, x^2+15x-13x-195=0

or, x(x+15)-13(x+15)=0

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

Either, x-13=0 Or, x+15=0

x=13 x=-15

Integers can be both positive or negative.

When x=13

Then 1st integer will be 13 and 2nd integer will be 13+2=15

When x=-15

Then 1st integer will be -15 and 2nd integer will be (-15+2)=-13

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the 1st required consecutive odd numbers be x

And the 2nd required number be (x + 2).

Then,

According to the Question,

⇒ x² + (x + 2)² = 394

⇒ 2x² + 4x - 390 = 0

⇒ x² + 2x - 195 = 0

⇒ x² + 15x - 13x - 195 = 0

⇒ x(x + 15) - 13(x + 15) = 0

⇒ (x + 15) (x - 13) = 0

⇒ x + 15 = 0 or x - 13 = 0

⇒ x = - 15, 13 (As x can't be negative)

⇒ x = 13

1st Number = x = 13

2nd number = x + 2 = 13 + 2 = 15

Hence, the required numbers are 13 and 15.