Question

The sum of the square of two consecutive odd positive integers is 290 . Find them

Answer 1

Let the first no. be a
Second No. = a+2

According to question
a^2 + (a+2)^2 = 290
=> a^2 + a^2 + 4a+ 4 = 290
=> 2a^2 + 4a - 286 = 0
=> a^2 + 2a - 143 = 0
=> a^2 +13a-11a-143 = 0
=> a(a+13) - 11(a+13) = 0
=> (a-11)(a+13) = 0

a=11 and - 13

Neglecting negative value, we get
a=11
First no. = 11
Second no. = 13

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