Word problems on Quadratic Equations (7)

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 = 181
Therefore,

`(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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