#### Question

Find two consecutive natural numbers such that the sum of their squares is 181.

#### Solution

Let the one number be x, therefore the other number is x + 1.#### Given

The sum of the squares of numbers = 181Therefore,

`(x)^2 + (x + 1)^2 =181`

Applying the formula `(a + b)^2 = (a^2 + 2ab + b^2)`

`x^2 + (x + 1)(x + 1) = 181`

`x^2 + x^2 + x + x + 1 = 181`

`2x^2 + 2x + 1 = 181`

Taking constant 181 to LHS

`2x^2 + 2x + 1 -181= 0`

`2x^2 + 2x - 180= 0`

Splitting the middle term

`2x^2 + 20x - 18x -180= 0`

Factoring the first two terms and the second two terms

`2x(x + 10) - 18(x + 10) = 0`

Factoring out the common x + 10

`(2x - 18) (x + 10) = 0`

By zero product rule

`2x - 18 = 0` or `x + 10 = 0`

`2x = 18` `x = -10`

`x = 9` and `x = -10`

`x = {9 , -10}`

Out of the two solutions 9 and -10, we are going to take only 9 as one of our answers, because the question states that there are two consecutive natural numbers (Natural numbers are all positive integers except 0). So one of the natural numbers is 9 and the other one is 9 + 1 = 10 (consecutive means one more than).

Answer is `x = 9 and 10`

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