the product of two consecutive even numbers is 12 more than the square of the smallest number.Find the numbers
x(x+2)=12+x^2
Let's assume the two consecutive even numbers are x and x+2.
According to the given condition, the product of these two numbers is 12 more than the square of the smallest number:
(x)(x + 2) = x^2 + 12
Expanding the left side of the equation:
x^2 + 2x = x^2 + 12
Simplifying the equation:
2x = 12
Dividing both sides by 2:
x = 6
So, the smallest number is 6.
The next consecutive even number would be:
x+2 = 6+2 = 8
Therefore, the two numbers are 6 and 8.
Let's represent the two consecutive even numbers as x and x+2 (with x being the smaller number). According to the problem, the product of these two numbers is equal to 12 more than the square of the smallest number:
x(x+2) = x^2 + 12
To solve this equation, we need to simplify and bring all the terms to one side:
x^2 + 2x = x^2 + 12
Now, subtracting x^2 from both sides, we have:
2x = 12
Dividing both sides of the equation by 2, we find:
x = 6
Therefore, the smaller number is 6. To find the larger number, we add 2 to it:
x+2 = 6+2 = 8
So, the two consecutive even numbers are 6 and 8.