Question

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

Answer 1

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Answer 2

Let the two odd positive integers are x+1, x+3

According to question :-

=> (x+1)² + (x+3)² = 290
=> x² + 2x + 1 +x² + 6x + 9 = 290
=> 2x² + 8x + 10 - 290 = 0
=> 2x² + 8x - 280 = 0
=> x² + 4x - 140 = 0
=> x² + (14-10)x - 140 = 0
=> x² + 14x -10x - 140 = 0
=> x(x+14) -10(x+14) = 0
=> (x+14)(x-10) = 0

Now, x = -14 and x = 10
from here, x = -14 is rejected
Hence, x = 10

Thus, the odd positive integers are = 11 and 13


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