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

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9 and 7

11 and 13

5 and 7

1 and 3

answer is B.

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Detailed Solution

Suppose one of the even positive numbers = x.
Other odd positive integer = x+2
Now sum of their square is 290. So,

290 = x^{2} + (x + 2)^{2}

290 = x^{2} + x^{2} + 4 x + 4

290 = 2 x^{2} + 4 x + 4

2 x^{2} + 4 x = 286

2 x^{2} + 4 x - 286 = 0

x^{2} + 2 x - 143 = 0

x^{2} + 13 x - 11 x - 143 = 0

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

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

Thus,

x = 11 or - 13

Neglect negative value..
Here,

x = 11

and

(x + 2) = 11 + 2 = 13

.
So, the odd positive integers are 11 and 13.
Correct option is 2.

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