find two consecutive odd positive integers sum of whose squares is 290.

Dear student

Let the two consecutive odd number be x and x + 2.

Given: Sum of the squares of two consecutive odd positive integers is 290.

 

According to question,

 

⇒(x)2+(x + 2)2= 290

⇒x2 + x2 + 4x + 4 = 290

⇒2x2+ 4x + 4 = 290

⇒2x2+ 4x = 290 - 4

⇒2x2 + 4x = 286

⇒x2+ 2x = 143

⇒x2 + 2x - 143 = 0

⇒x2 + 13x - 11x -143=0

⇒x(x + 13) -11(x + 13 )=0

 

⇒x = 11 (since they r positive integers)

⇒x + 2 = 11+2=13

 

 

Thus, the two consecutive odd positive integers are 11 and 13.


Regards

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