The sum of the squares of three consecutive positive numbers is 365. W...
Problem Analysis
We are given that the sum of the squares of three consecutive positive numbers is 365. Let's assume the three consecutive numbers as x, x+1, and x+2.
Solution
We can represent the sum of the squares of these three numbers as an equation:
x^2 + (x+1)^2 + (x+2)^2 = 365
Expanding the equation:
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 365
3x^2 + 6x + 5 = 365
3x^2 + 6x - 360 = 0
Divide the equation by 3:
x^2 + 2x - 120 = 0
Factorizing the Equation
We need to factorize the quadratic equation x^2 + 2x - 120 = 0 to find the values of x.
(x + 12)(x - 10) = 0
From the equation, we have two possible values for x:
x + 12 = 0 or x - 10 = 0
If x + 12 = 0, then x = -12, which is not a positive number. Hence, we discard this solution.
If x - 10 = 0, then x = 10. This gives us the first number as 10.
Calculating the Other Numbers
Using the value of x = 10, we can calculate the other two consecutive numbers:
First number: x = 10
Second number: x + 1 = 10 + 1 = 11
Third number: x + 2 = 10 + 2 = 12
Sum of the Numbers
The sum of the three numbers is:
10 + 11 + 12 = 33
Therefore, the correct answer is option 'B', which is 33.