find four consecutive integers such that the sum of the squares of the first two is 11 less than the square of the fourth
Let x, x+1, x+2, x+3 be the four consecutive integers.
"the sum of the squares of the first two is 11 less than the square of the fourth"
x^2 + (x+1)^2 + 11 = (x+3)^2
x^2 + x^2 + 2x + 1 + 11 = x^2 + 6x + 9
2x^2 + 2x + 12 = x^2 + 6x + 9
x^2 - 4x + 3 = 0
(x-3)(x-1) = 0
x = 1, 3
So the four consecutive integers are 1, 2, 3, 4; or 3, 4, 5, 6.
Let's represent the four consecutive integers as x, x+1, x+2, and x+3.
We're given that the sum of the squares of the first two integers is 11 less than the square of the fourth integer. Mathematically, we can express this as:
x^2 + (x+1)^2 = (x+3)^2 - 11
Expanding and simplifying the equation:
x^2 + x^2 + 2x + 1 = x^2 + 6x + 9 - 11
Combining like terms:
2x^2 + 2x + 1 = x^2 + 6x - 2
Rearranging the equation:
2x^2 - x^2 + 2x - 6x = -2 - 1
x^2 - 4x = -3
Bringing all terms to one side:
x^2 - 4x + 3 = 0
Factoring the equation:
(x - 3)(x - 1) = 0
Setting each factor to zero:
x - 3 = 0 or x - 1 = 0
Solving for x:
x = 3 or x = 1
Therefore, the four consecutive integers can be x = 3, 4, 5, 6 or x = 1, 2, 3, 4.
To find the four consecutive integers, let's introduce variables. Let's represent the first integer as x, so the second consecutive integer will be x + 1, the third will be x + 2, and the fourth will be x + 3.
According to the problem, the sum of the squares of the first two integers is 11 less than the square of the fourth integer. Formulating this in an equation:
(x^2 + (x + 1)^2) = (x + 3)^2 - 11
Now, let's solve this equation step-by-step:
Expand the equation:
x^2 + x^2 + 2x + 1 = x^2 + 6x + 9 - 11
Combine like terms:
2x^2 + 2x + 1 = x^2 + 6x - 2
Rearrange the equation to solve for x:
2x^2 - x^2 + 2x - 6x = -2 - 1
Simplify:
x^2 - 4x = -3
Move all terms to one side:
x^2 - 4x + 3 = 0
Factorize the quadratic equation:
(x - 1)(x - 3) = 0
Setting each factor to zero, we get:
x - 1 = 0 or x - 3 = 0
Solve for x:
x = 1 or x = 3
Thus, the two possible values for x are 1 and 3.
For x = 1:
The four consecutive integers are 1, 2, 3, and 4.
For x = 3:
The four consecutive integers are 3, 4, 5, and 6.
Therefore, there are two possible sets of four consecutive integers that satisfy the given conditions: {1, 2, 3, 4} and {3, 4, 5, 6}.