# 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

⇒x^{2 }+ x^{2 }+ 4x + 4 = 290

⇒2x^{2}+ 4x + 4 = 290

⇒2x^{2}+ 4x = 290 - 4

⇒2x^{2} + 4x = 286

⇒x^{2}+ 2x = 143

⇒x^{2 }+ 2x - 143 = 0

⇒x^{2} + 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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