Please Help Me
The Sum Of Two Square Of Two Consecutive Natural Number Is 313. Find The Numbers.
let the two numbers be n and n+1
then:
n^2 + (n+1)^2 = 313
2n^2 + 2n - 312 = 0
n^2 + n - 156 = 0
(n+13)(n-12) = 0
n = 12 or n + -13, but n> 0
the numbers are 12 and 13
To find the two consecutive natural numbers whose sum of their squares is 313, we can break down the problem into equations.
Let's assume the first number is "x" and the second number is "x + 1" since they are consecutive natural numbers.
According to the problem, the sum of their squares is 313. So we can write the equation as follows:
x^2 + (x + 1)^2 = 313
Now we need to solve this equation to find the values of x and x + 1.
Expanding the equation, we get:
x^2 + (x^2 + 2x + 1) = 313
Combining like terms, we have:
2x^2 + 2x + 1 = 313
Rearranging the equation, we obtain:
2x^2 + 2x - 312 = 0
Now we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 2, b = 2, and c = -312.
Plugging in these values, the formula becomes:
x = (-2 ± √(2^2 - 4 * 2 * -312)) / (2 * 2)
Simplifying the formula further, we get:
x = (-2 ± √(4 + 2496)) / 4
x = (-2 ± √2500) / 4
Now we calculate the square root separately:
√2500 = 50
So our equation becomes:
x = (-2 ± 50) / 4
Now we substitute the positive and negative values for x:
For the positive value:
x = (-2 + 50) / 4
x = 48 / 4
x = 12
So the first number is 12.
For the negative value:
x = (-2 - 50) / 4
x = -52 / 4
x = -13
Since natural numbers cannot be negative, the value of x = -13 doesn't satisfy the problem's condition.
Thus, the two consecutive natural numbers whose sum of their squares is 313 are 12 and 12 + 1 = 13.