Here's the question you clicked on:
diamondeye10
the product of two consecutive positive even numbers is 1520. What are the numbers? when I try "n" and "n+2" i get an odd number
you can express those consecutive even numbers like "n" and "n+2" and their product is (n)(n+2) = 1520
guess and check for this one there is no other way
\(40\times 42=1680\) nope, too big. try again with smaller numbers
That's not quite the right equation...
@Albert0898 it is the PRODUCT not the sum
you cannot use an equation to solve this, only to rewrite the question, don't be fooled you just have to grind it until you find it
satellite was close ;)
hahah, sorry i used your guess to approximate.
yeah a calculator helps i started too high, went down a couple, got it in try two
n(n+2) = 1520 n^2 + 2n - 1520 = 0 (n-38)(n+40) = 0 Since the numbers are positive, the 2nd solution doesn't work. n must be 38 and n + 2 is 40.
(I couldn't factor that as easily as I made it sound... so a calculator helped me too :) )
You definitely do not have to grind it out and guess. But the equation is not shown correctly above. It is (2N)(2N + 2) = 1520. Becomes a simple quadratic. Solve for N. Answer will be the 2N and 2N + 2.
so my question is, how did you know how to factor ? you had to come up with two numbers whose product is 1520 and that are two apart in other words just rewrote the question, still had to guess and check
x = even number x + 2 = consecutive even number x + 2 * x = 1520 x + 2 = 1520/x x = 1518/x x * x = 1518 Find the square root of 1518. Approximate and Closest Even Number: 38. x = 38 x + 2 = 40
i guess if you wanted to be ridiculous you could write \[x^2+2x=1520\] and complete the square \[(x+1)^2=1521\] \[x+1=\sqrt{1521}=39\]\[x=38\] but that is a silly thing to do
@satellite73 Yes, I agree with you... that's what I meant... you either guess and check to factor, or just guess and check vs. 1520. My only point was that it is possible to write an equation to solve something like this.
@JakeV8 yes you are right, but i maintain that changing a problem like this in to an equation and then solving it by solving the original problem in words is a dog chasing its tail
You guys are missing it all together. Set the variables as 2N and 2N + 2. If you set N to a positive integer, 2N and 2N + 2 are automatically even. Becomes a simple quadratic. See my previous post.
thanks all I think I found my problem
Ok, I'm here. Give me a second to refresh myself on this.
Ok, I'm up to speed. I still stand firmly on my answer methodology. The way to GUARANTEE that you have two even numbers is to use (2N)(2N + 2) = 1520 because 2N and 2N + 2 will be FORCED to be even if N is a positive integer. Absolutely. I strongly suggest you use this methodology. If I understand, your current question has something to do with 3 positive even numbers now? Is that correct?
And please, whatever you do, don't follow satellite73 on this problem. He is WAY off base suggesting that you guess. That's beyond ludicrous.
@tcarroll010, that is the equation I was using however, i am still unable to find two positive even numbers. No I do not have to find three evennumbers, I was seeing if the way I was working it would give even numbers for three. hope that makes sence, sorry my computer is slow. . I only need two positive even numbers