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Author Topic: Hardest question in the world  (Read 520 times)
Fujisawa4654
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« on: February 03, 2004, 06:18:07 PM »

According to me, that is...

What 2 numbers when added equal 10 and when multiplied equal 130?

Think Algebra 2...
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theravenisdead
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« Reply #1 on: February 03, 2004, 07:06:11 PM »

I would love to answer this intriguing question, but alas I am an idiot  -_-
« Last Edit: February 03, 2004, 07:06:52 PM by theravenisdead » Logged
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« Reply #2 on: February 04, 2004, 08:23:24 AM »

Quote
According to me, that is...

What 2 numbers when added equal 10 and when multiplied equal 130?

Think Algebra 2...

Hrmph. Unless this is some kinda trick question, I don't think this has a tidy answer. I might have done the math wrong, though... Observe:

1. So, x+y=10, and xy=130.

2. Then y=130/x... so, x+(130/x)=10.

3. Multiply both sides by x, and x^2+130=10x.

4. Subtract 10x from both sides, and x^2-10x+130=0.

5. Now for the Quadratic Equasion! x=(-b+-sqrt(b^2-4ac))/2a.
So, x=(-(-10)+-sqrt((-10)^2-(4*1*130)))/(2*1)

6. -(-10) = 10, and those 1's are useless, so let's fix that:
x=(10+-sqrt((-10)^2-(4*130)))/2

7. (-10)^2=100. So, x=(10+-sqrt(100-(4*130)))/2

8. 4*130=520. So, x=(10+-sqrt(100-520))/2

9. 100-520=-420. So, x=(10+-sqrt(-420))/2

10. Uh, oh... Square root of a negative number. That means our results are going to be imaginary.
sqrt(-1)=i, so let's pull that out.
x=(10+-(i*sqrt(420)))/2

11. sqrt(420)=sqrt(4)*sqrt(105).
So, x=(10+-(i*sqrt(4)*sqrt(105)))/2
This may seem pointless, but wait for it...

12. sqrt(4)=2. So x=(10+-(2i*sqrt(105)))/2

13. Now, divide both the top and bottom by 2.
So, x= ((10/2)+-((2i*sqrt(105)/2)))/(2/2)

14. (2i*sqrt(105))/2=i*sqrt(105), 10/2=5, and 2/2=1.
So, x=(5+-i*sqrt(105))/1

15. And anything divided by 1 is itself.
So, x=5+-i*sqrt(105)

16. That leaves us with two possible values for x:
5+i*sqrt(105) and 5-i*sqrt(105)

17. Solve for y:
10-x=y. So 10-(5+i*sqrt(105))=y. So, y=5-i*sqrt(105). Or
10-x=y. So 10-(5-i*sqrt(105))=y. So, y=5+i*sqrt(105).

18. And there's our answers. Either

x=5+i*sqrt(105) and y=5-i*sqrt(105)

or

x=5-i*sqrt(105) and y=5+i*sqrt(105)

Geh. Imaginary numbers. Gotta love 'em. :P Any questions, class?
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Fujisawa4654
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« Reply #3 on: February 04, 2004, 10:04:17 AM »

 :bawl :bawl :bawl :bawl

Nobody could answer it on the other forums!!
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« Reply #4 on: February 04, 2004, 11:39:31 AM »

Quote
:bawl :bawl :bawl :bawl

Nobody could answer it on the other forums!!

*Shrugs* It's a tough little piece of math. I think the answer I gave you is correct, though: The two numbers are 5+i*sqrt(105) and 5-i*sqrt(105)

I was pretty sure that there weren't going to be any nice, real number answers to this, cause for two numbers to add up to 10, they either need to

a. Both be non-negative and less than 10, or
b. Exactly one of them must be negative (or 0).

We know that one of them can't be negative (or 0), 'cause when you multiply a negative number by a positive number, the result is always negative (and any number multiplied by 0 is 0), and these two numbers when multiplied become a positive number: 130.

Unfortunately, if both of them are non-negative and less than 10, then the most that they could multiply up to is 25: 5 times 5. That's FAR short of 130.

So, to find the messy answer we were doomed to get, we have to combine the two equasions to get ourselves down to just one variable, apply the Quadratic equasion, and hope for the best.

Are you studying imaginary numbers and/or the quadratic equasion in class at the moment? If so, I think this answer is correct. If this is stuff you haven't covered, then I'm guessing this was either a question that was intended to stump you (because you can't figure it out unless you know about imaginary numbers), or the question was mis-written. If 130 was 13 instead, for instance, it would have a nice, tidy real number answer.

Yay, my Mathematics Minor is actually coming in handy! ^_^V
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« Reply #5 on: February 04, 2004, 11:56:39 AM »

For my particular field of work [i.e. hobby], I don't have use for such ugly equations.  In fact, since most of my stuff is all about kicking performance out of computer systems, I typically don't want to even do multiplication if I can help it.  :P
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Dub vs. Sub, let's keep quiet about it.
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« Reply #6 on: February 04, 2004, 04:24:20 PM »

Not even here am I safe from the vile reach of Algebra...

::shoots self::
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