Question

find the consective positive odd numbers whose squares sum is 394

Answer 1

Hi Mate !!



Let the first consecutive odd integer be x and second be ( x + 2 )


• Sum of the square is 394

x² + ( x + 2 )² = 394

[ Using identity :- ( a + b )² = a² + 2ab + b² ]

x² + x² + 4x + 4 = 394

2x² + 4x + 4 - 394 = 0

2x² + 4x - 390 = 0

[ Dividing both side by 2 ]

x² + 2x - 195 = 0

Now , factorising above equation !!


x² + 15x - 13x - 195 = 0

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

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

° ( x + 15 ) = 0

x = -15 ...........( neglected bcoz, it is given in question that the nos. are positive odd )


° ( x - 13 ) = 0

x = 13


So, 1st no. is x = 13
2nd no.is ( x + 2 ) = 15

And hence , 13 and 15 are consecutive odd positive integer having square sum as 394 !!

Answer 2

Answer:

Step-by-step explanation:

Let the first consecutive odd integer be x and second be ( x + 2 )

• Sum of the square is 394

x² + ( x + 2 )² = 394

[ Using identity :- ( a + b )² = a² + 2ab + b² ]

x² + x² + 4x + 4 = 394

2x² + 4x + 4 - 394 = 0

2x² + 4x - 390 = 0

[ Dividing both side by 2 ]

x² + 2x - 195 = 0

Now , factorising above equation !!

x² + 15x - 13x - 195 = 0

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

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

( x + 15 ) = 0

x = -15 ( neglected because it is given in question that the nos. are positive odd )

( x - 13 ) = 0

x = 13

So, 1st no. is x = 13

2nd no.is ( x + 2 ) = 15

And hence , 13 and 15 are consecutive odd positive integer having square sum as 394