how to find two consecutive positive even integers such that the square of first increased by 2 is equal to 3 times the second.
If the even numbers are x and x+2, then you need
x^2 + 2 = 3x
Solve for x, and then discard the unwanted solution.
To find two consecutive positive even integers that satisfy the given condition, we can follow these steps:
Step 1: Assume the first even integer as "x".
Step 2: Since the second even integer is consecutive, we can represent it as "x + 2" (as it will be 2 greater than the first even integer).
Step 3: According to the given condition, the square of the first even integer increased by 2 is equal to 3 times the second even integer. This can be written as:
x^2 + 2 = 3(x + 2)
Now we can solve this equation to find the value of "x" and subsequently the two consecutive even integers.
Step 4: Simplify the equation by expanding and rearranging terms:
x^2 + 2 = 3x + 6
x^2 - 3x - 4 = 0
Step 5: Factorize the quadratic equation:
(x - 4)(x + 1) = 0
Step 6: Apply the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Set each factor equal to zero and solve for "x":
x - 4 = 0 or x + 1 = 0
x = 4 or x = -1
Step 7: Since we are looking for positive even integers, we choose the value of "x" as 4.
Step 8: Substitute the value of "x" back into the expression for the two consecutive even integers:
First even integer = x = 4
Second even integer = x + 2 = 4 + 2 = 6
Therefore, the two consecutive positive even integers that satisfy the given condition are 4 and 6.