The product of two consecutive integers, n and n+1, is 42. What is the positive integer that satisfies the situation? Is it 6 x 7?
But, did you solve it using algebra?
n(n + 1) = 42
n^2 + n - 42 = 0
( n-6)(n+7) = 0
n = 6 or n = -7
n cannot = -7
therefore n = 6 and n + 1 = 7
Your answer is correct. How did you get it?
That two consecutive numbers that multiply and equal 42 is 6x7. 6 + 1+7
Well, if we take n as the first integer, and n+1 as the next consecutive integer, we can rewrite the situation as n(n+1) = 42. Now let's solve this using the quadratic formula, the square root of holy math! Just kidding, the quadratic formula is:
(-b ± √(b^2 - 4ac)) / 2a
But in this case, we have n and n+1 as our variables, so it becomes:
(-(n+1) ± √((n+1)^2 - 4n)) / 2n
Now, when we calculate this equation to find the possible values of n, I must tell you that this is going to be a serious calculation, not a clownish one.
Using this formula, we find that n can be either -6 or 7. Since we are looking for positive integers, the answer is 7, not -6. So, no clown-juggling needed here! The positive integer that satisfies n(n+1) = 42 is indeed 7.
To find the positive integer that satisfies the situation, we need to find two consecutive integers whose product is 42.
Let's set up an equation:
n * (n+1) = 42
We can solve this equation by trying different values for n.
First, let's try n = 6:
6 * (6+1) = 6 * 7 = 42
So, when n = 6, the product of the two consecutive integers is indeed 42.
Therefore, the positive integer that satisfies the situation is 6.