Find two consecutive of the odd natural numbers in the sum of the whose squares is 202

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Let the two consecutive odd numbers be x and x+2 .

It is given that the sum of their squares is equal to 202. Therefore we have x^2 + (x+2)^2 = 202

=> x^2 + x^2 + 4x + 4 = 202

=> 2x^2 + 4x + 4 = 202

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Let the two consecutive odd numbers be x and x+2 .

It is given that the sum of their squares is equal to 202. Therefore we have x^2 + (x+2)^2 = 202

=> x^2 + x^2 + 4x + 4 = 202

=> 2x^2 + 4x + 4 = 202

subtract 202 from both the sides

=> 2x^2 + 4x - 198 =0

divide all terms by 2

=> x^2 + 2x - 99 = 0

=> x^2 + 11x - 9x - 99 =0

=> x(x+11) -9(x+11) =0

=> (x -9) (x+11) =0

From x -9 =0 we get x = 9

and from x +11 = 0 we get x = -11 which is rejected.

Therefore the numbers are ( 9, 11).

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Let the first odd number be x

Then the next odd number is (x+2)

Given that the sum of the squares of the numbers is 202.

Then we will write the expression as follows.

x^2 + (x+2)^2 = 202

We will open brackets.

==> x^2 + x^2  + 4x + 4 = 202

Combine like terms.

==> 2x^2 + 4x + 4 = 202

Divide by 2:

==> x^2 + 2x + 2 = 101

==> x^2 + 2x - 99 = 0

==> ( x +11)( x- 9) = 0

==> Then, the numbers are : 11 and 9

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