Question

The sum of the squares of two consecutive odd positive integers is 202. find the integers

Answer 1

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Define x :
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Let one of the number be x. The other number is x + 2.

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Construction Equation :
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The sum of the square of the two number is 202.

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

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Solve for x :
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`x^2 + (x+2)^2 = 202`


`x^2 + x^2 + 4x + 4 = 202`


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


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


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


`(x+11)= 0 \ \ or \ \ (x-9) =0`


`x = - 11 \ or \ x = 9`

x = -11 is rejected, since it only want positive integer.

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Find the two numbers :
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x = 9.
x + 3 = 11

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Answer: The numbers are 9 and 11.
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