# Word problem/Quadratic equations

1. May 4, 2006

### Richay

Find 3 consecutive intergers such that the product of the first and the second is equal to the product of -6... and the third.
Answer in the from: a, b, c and d, e, f (smallest to largest)

So far, I haven't lifted off anywhere from this problem.

I DO understand intergers, i've passed my consecutive intergers test. But I still fail to solve this problem. I can't breakdown this problem. Can anybody give me an equation?

2. May 4, 2006

### Laceylb

hmm whenever i try it, i get really weird answers

3. May 4, 2006

### Laceylb

ok i got it..... x(x+1) = -6(x+2)

x^2 + x = -6x -12

x^2+7x+12=0

(x+3)(x+4)=0

x=-3 and -4

4. May 4, 2006

### Richay

What i dont get it..

If there's 3 intergers, why do i have to answer with 6?

5. May 5, 2006

### Laceylb

well i think because there are two different answers for x and when you plug it back in you would have 6 total numbers, im just taking a guess as that

6. May 5, 2006

### VietDao29

Laceylb, you seem to be new here. Here, we don't show complete answer to the OP, we can however guide them to the correct answer. And you should only give him / her full answer when you see that the OP has tried his / her best but for somehow cannot solve the problem. The rules can be viewed here.
IMHO, giving complete answer without any explanation is not a good way of learning and teaching. :)

7. May 5, 2006

### Laceylb

im sorry, i am new, but i didnt think i solved it all, he had to put the x's back into the problem

8. May 5, 2006

### GregA

Richay...try to identify the range of numbers that the solution definitely *isn't*.
For example...say I let n+2 = 4...what is the product of 4 and -6?...what is the product of 2 and 3? ( since n+2 = 4, our third number...our first and second are n and n+1 )...
secondly can n be somewhere around +/- 10 million?...whereabouts does n lie?

Be aware that potentially the problem might be expressed in any of these ways:
-6n = (n+1)(n+2)
-6n = (n-1)(n-2)
-6n(n-2) = (n-1)(n)
-6n(n+2) = (n+1)(n)

Only two of these is any help, the other two will not...which two cannot help and why?
(By finding the range of numbers that your solution isnt, you should be able to solve it just by inspection and then you should try to justify your choices by choosing and solving the correct quadratics)

Last edited: May 5, 2006
9. May 5, 2006

### Richay

I see, but that makes it confusing for me. That's why i don't understand what to do.

I'll take a shot at this.
-6n = (n+1)(n+2)...so-so?? I can see that this problem can be figured out, but I can't find the correct numbers... HELP ME. Omg, i can't find a number that will fit in with all the problems. I think n is around 8 in this one. But i can't figure it out.

-6n = (n-1)(n-2)...Works. Because:
-6 x -2 = 12 (-3) x (-4) = 12. (Lacey showed me this solution eariler)

-6n(n-2) = (n-1)(n) This can't possibly work. You can always prove me wrong though if you like. But you will never even get a close answer because the first equation will have a HIGH total. And the second equation will always be lower. SO, screw this one for now.

-6n(n+2) = (n+1)(n) Screw this one too.

Well, that's as far as i got. What do i need to do next to progress?

10. May 6, 2006

### GregA

*Edit* Major sorry ...I noticed that I typoed in my first post. The following two below include an extra n

"6n(n-2) = (n-1)(n) This can't possibly work. You can always prove me wrong though if you like." No need (what what this does is multiply -6 by the first and last: -6n(n+2) and make it equal to the product of the second and third n(n-1))

"-6n(n+2) = (n+1)(n) Screw this one too." again my apologies (similar reason)

Luckily, in this problem if you know whereabouts n lies you haven't got many choices and can just throw a few numbers at it until you hit the solution...It is still better if you actually solve one of the 2 correct quadratics and actually make sense of what results they yield.

take -6n = (n+1)(n+2)...this can be manipulated to give n^2 +9n + 2 =0...why does this quadratic not help?...well most importantly its roots are not integers. the reason?
For integer values of 1 >= n >= - 8... -6n > (n+1)(n+2) and once n becomes -9 or smaller (or n becomes zero or positive) then -6n < (n+1)(n+2). Here our numbers are in descending order (n < n+1) This result suggests that we should try letting our numbers should be in ascending order instead.

Last edited: May 6, 2006
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