Question

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

Answer 1

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$Let \: the \: two \: consecutive \: odd \: numbers \: be \: 'x'\: and \: 'x + 2' \\ \\ Given \: that \: sum \: of \: square \: of \: two \: consecutive \: odd \: numbers \: is \: 290 \\ \\ Now, \: Acc \: to \: statement : \\ \\ = > x {}^{2} + (x + 2) {}^{2} = 290 \\ \\ = > x {}^{2} + x {}^{2} + 4 + 4x = 290 \\ \\ = > 2x {}^{2} + 4x + 4 = 290 \\ \\ = > 2 x {}^{2} + 4x - 286 = 0 \\ \\ = > 2(x { }^{2} + 2x - 143) = 0 \\ \\ = > x {}^{2} + 13x - 11x - 143 = 0 \\ \\ = > x(x + 13) - 11(x + 13) = 0 \\ \\ = > x = - 13 \: and \: x = 11$

So , The first consecutive odd positive number is 11 and the Second con. odd positive number is 13...
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Answer 2

Solution :

Let x , ( x + 2 ) are consecutive odd

numbers ,

according to the problem given ,

Sum of the squares of given

numbers = 290

=> x² + ( x + 2 )² = 290

=> x² + x² + 2² + 4x - 290 = 0

=> 2x² + 4x - 286 = 0

divide each term with 2 , we get

=> x² + 2x - 143 = 0

Splitting the middle term ,

=> x² + 13x - 11x - 143 = 0

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

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

Therefore ,

x + 13 = 0 or x - 11 = 0

x = -13 or x = 11

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