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

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An Intiative by Sri Chaitanya

11, 33

11, 53

11, 13

21, 13

answer is C.

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Detailed Solution

Given two consecutive positive integers, sum of squares is 290.
Let one of the odd positive integers be x.
Then the other odd positive integer be

x + 2

.
The sum of square

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

\Rightarrow x^{2} + x^{2} + 4 x + 4

\Rightarrow 2 x^{2} + 4 x + 4

Given that their sum of squares

= 290

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

\Rightarrow 286 = 2 x^{2} + 4 x

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

Now, factorising the equation,

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

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

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

\Rightarrow (x + 13) = 0 or (x - 11) = 0

\Rightarrow x = - 13 or x = 11

We have to consider positive integer, so, x=11.
Thus, the other number is x+2=11+2=13.
So, value of odd integer is

11, 13

.
Therefore, option (3) is correct.

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