Question

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

Answer 1

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Answer 2

SOLUTION

Let x be an odd positive Integer.

Then next odd positive integer= x+2

As per given, we have

$= > {x}^{2} + (x + 2) {}^{2} = 290 \\ = > 2x {}^{2} + 4x - 286 = 0 \\ = > {x}^{2} + 2x - 143 = 0 \\ = > {x}^{2} + 13x - 11x - 143 = 0 \\ = > x(x + 13) - 11(x + 13) = 0 \\ = > (x + 13)(x - 11) = 0 \\ \\ = > x = - 13 \: \: and \: \: \: x = 11$

Since x> 0 therefore x= 11

Hence Required integer are 11 and 13.

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