The sum of the squares of three consecutive positive integers is 365
What are these integers?
Please write this problem in quadratic equation form please
let x = smallest integer, then the other integers are x+1 and x +2.
x^2 + (x+1)^2 + (x+2)^2 = 365
Square terms and combine.
X^2+(x+1^2)+(x+2^2)=365 3x^2+6x-360=0 x^2+2x-120=0 (x+12)(x-10)=0 x=10 or -12 Answer=10,11,12 or -12,-11,-10
Let's represent the three consecutive positive integers as x, (x+1), and (x+2).
According to the problem, the sum of the squares of these integers is 365. We can write this in equation form as:
x^2 + (x+1)^2 + (x+2)^2 = 365
Now let's simplify and solve the quadratic equation.
To find the three consecutive positive integers, let's assume the first integer as "x".
The next two consecutive positive integers would be (x + 1) and (x + 2) respectively.
The sum of their squares can be expressed as:
x^2 + (x + 1)^2 + (x + 2)^2 = 365
Expanding and simplifying the equation, we get:
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 365
3x^2 + 6x + 5 = 365
Now let's rearrange the equation in quadratic form by moving 365 to the right side:
3x^2 + 6x + 5 - 365 = 0
3x^2 + 6x - 360 = 0
Hence, the equation in quadratic form representing the problem is:
3x^2 + 6x - 360 = 0
Now, to find the values of x, you can solve this quadratic equation using factoring, completing the square, or using the quadratic formula.