the sum of the squares of two consecutive numbers is 265. find the numbers
n^2 + (n+1)^2 = 265
Solve for n.
Let's assume the two consecutive numbers are x and x + 1.
According to the problem, the sum of the squares of these two numbers is 265.
So, we can write the equation as:
x^2 + (x + 1)^2 = 265
Expanding and simplifying the equation:
x^2 + x^2 + 2x + 1 = 265
2x^2 + 2x + 1 = 265
2x^2 + 2x - 264 = 0
Dividing both sides of the equation by 2:
x^2 + x - 132 = 0
Now, we can solve the quadratic equation by factoring or using the quadratic formula.
Factoring the equation:
(x + 12)(x - 11) = 0
Setting each factor equal to zero:
x + 12 = 0 or x - 11 = 0
Solving for x:
x = -12 or x = 11
Since we need the consecutive numbers, we take the positive value:
x = 11
Therefore, the two consecutive numbers are 11 and 12.
To find the two consecutive numbers whose sum of squares is 265, let's set up the equation.
Let's assume the first number is "x". Since the numbers are consecutive, the second number will be "x + 1".
According to the problem, the sum of the squares of these numbers is 265. Mathematically, this can be expressed as:
x^2 + (x + 1)^2 = 265
Now, let's solve this equation step-by-step:
Expand the equation:
x^2 + (x^2 + 2x + 1) = 265
Combine like terms:
2x^2 + 2x + 1 = 265
Rearrange and simplify:
2x^2 + 2x + 1 - 265 = 0
2x^2 + 2x - 264 = 0
Divide the equation by 2 to simplify:
x^2 + x - 132 = 0
Now, we need to factorize this quadratic equation. We are looking for two numbers that multiply to -132 and add up to 1. The numbers are 12 and -11:
(x + 12)(x - 11) = 0
Now set each factor equal to zero and solve for x:
x + 12 = 0 ---> x = -12
x - 11 = 0 ---> x = 11
The two possible solutions are x = -12 and x = 11. However, since the problem states that the numbers are consecutive, we can discard the negative solution. Thus, the first number is 11, and the second number is 11 + 1 = 12.
Therefore, the two consecutive numbers whose sum of squares is 265 are 11 and 12.