The product of two consecutive positive even numbers is 728. What are the numbers? (Enter solutions from smallest to largest.)
and
What is the sum of the first 60 consecutive odd numbers?
To find the consecutive positive even numbers whose product is 728, we need to find two numbers that, when multiplied, equal 728.
Let's denote the smaller even number as x and the next consecutive even number as x + 2. Since they are consecutive even numbers, the next number will have a difference of 2 compared to the previous one.
According to the problem, their product is 728. We can set up the equation:
x(x + 2) = 728
Expanding this equation, we get:
x^2 + 2x = 728
Rearranging the equation, we have:
x^2 + 2x - 728 = 0
To solve this quadratic equation, we can either factorize it or use the quadratic formula. In this case, I will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = 1, b = 2, and c = -728. Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4*1*(-728))) / (2*1)
Simplifying further:
x = (-2 ± √(4 + 2912)) / 2
x = (-2 ± √(2916)) / 2
x = (-2 ± 54) / 2
Now, we have two cases to consider, one with the positive square root and one with the negative square root:
Case 1: Using the positive square root:
x = (-2 + 54) / 2
x = 52 / 2
x = 26
Case 2: Using the negative square root:
x = (-2 - 54) / 2
x = -56 / 2
x = -28
Since we are looking for positive even numbers, we discard the solution x = -28.
Therefore, the two consecutive positive even numbers whose product is 728 are 26 and 28.
The square root of 728 = 26.98
Take it from there.