## Pages

### 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