Example 11 : Find two consecutive odd positive integers, sum of whose squares
is 290.​

2 thoughts on “Example 11 : Find two consecutive odd positive integers, sum of whose squares<br />is 290.​”

1. Answer:

   Let x an odd positive integer

   Then, according to question

   x² + ( x + 2 ) = 290

   2x² + 4x – 286 = 0

   x² + 2x − 143=0

   x² + 13x − 11x − 143=0

   ( x +13 ) ( x + 11 ) = 0

   x = 11 as x is positive

   Hence required integers are 11 and 13.

   Step-by-step explanation:

   I hope it’s helpful for you.

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2. Answer:

   11 and 13

   Step-by-step explanation:

   11^2+13^2= 121 + 169 =290

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