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xddxogm3
02-04-2005, 09:48 PM
Me and another classmate are debating the true answer to the following equation. This is a college algebra question.

x^(3/2) + 8x^(1/2) + 16x^(-1/2) = 0

The book says no solution.
My classmate says it has a solution.

I feel the negative exponent will prevent it from being solved. Is this correct, or is there more to it?

If it can be solved, can you show me the proof.

<edit 1> sorry edited out his proof on accident. i will put it back in tomorrow. </edit 1>

JaWiB
02-04-2005, 10:14 PM
>>x^(-1/2) * x^(2) = x^(3/2)

Where did this come from? Yes, it's correct, but I don't see it relating to the original problem.

if you graph it you find that there is no solution at 5 or at any other point. Just try:

5^(3/2) + 8(5)^(1/2) + 16(5)^(-1/2) = 36.2243

Edit: I think I see what he's trying to do. You are trying to solve for zero, though:

x^(-1/2)x^2 = 0
x=0

But from the original problem you would put x=0 and find:

16x^(-1/2) = 16/x(^1/2) = 16/0 = undefined

SourceCode
02-04-2005, 11:18 PM
Your book is wrong.

x^(3/2) + 8x^(1/2) + 16x^(-1/2) = 0

has a solution at x = -4

Substituting -4 into the left side of equation we get

(-4)^(3/2) + 8*(-4)^(1/2) + 16*(-4)^(-1/2)
(2i)^3 + 8*2i + 16/(2i)
8i + 16i + 8/i
8i + -8i
0

note that 8/i = 1/i * 8 = -1 * 8 since 1/i = -1

xddxogm3
02-04-2005, 11:57 PM
i didn't think of graphing the function until now.
i graphed the original function
x^(3/2)+8x^(1/2)+16x^(-1/2)
and there is no zero that is formed when graphed.
please someone else provide your input to resolve this.

SourceCode

note that 8/i = 1/i * 8 = -1 * 8 since 1/i = -1
i saw what you were doing after a while. i'm assuming the -1 should be -i.
i=sqrt(-1)
i^2 = -1
1/i != -1 //even with manipulation of the denominator i get 1/i*i/i=-i
1/i = 1/sqrt(-1)
I could be wrong, i'm not a mathematician, but last i checked
sqrt(-1) = i
i^2 = -1
i^3 = -i
i^4 = 1

Sang-drax
02-05-2005, 03:48 AM
Every equation like this has solutions, but they might be complex.

However, this is a simple 2nd degree equation.

x^(3/2) + 8x^(1/2) + 16x^(-1/2) = 0

multiply with x^(1/2)

x^2 + 8x + 16 = 0

Now it's a normal 2nd degree equation with only one root and it is:

-4

So there are real roots, but to be able to put -4 back in the original equation, you'll need complex numbers, like SourceCode showed (although he calculated wrong, he came up with the correct answer).

EDIT:

i didn't think of graphing the function until now.
i graphed the original function
x^(3/2)+8x^(1/2)+16x^(-1/2)
and there is no zero that is formed when graphed.
please someone else provide your input to resolve this
Here's a plot of the function. When x<0 the function is complex-valued. The zero at x=-4 is clearly shown (Re y=Im y=0).

SourceCode
02-05-2005, 08:09 AM
1/i != -1 //even with manipulation of the denominator i get 1/i*i/i=-i

yea sorry typo, 1/i = -i

SourceCode
02-05-2005, 08:37 AM
And if you are confused why 1/i = -i, a simplified algebraic way of looking at it is

Knowing i^2 = -1, divide by sides by i and then
i^2/i = -1/i
i= -1/i
-i = 1/i

What clued me in into trying -4 was the same thing that clued Sang-Drax in I think, because like he said if you notice that multiplying x^(1/2) by both sides you end up with x^2 + 8x + 16 = (x + 4)^2 which has a double root at x = -4. I guess the reasoning behind the thought process is that when you see something like that you just notice it. It's just one of those things that just comes from seeing it done a couple of times. I think this is way too tough for a college algebra course, most people don't see stuff like this until much later. I didn't feel really comfortable working with lots of fractional exponents till I studied differential equations, lots of nasty algebraic manipulations there hehe.

Great post btw Sang, great thread in general!

xddxogm3
02-05-2005, 01:34 PM
JaWib, SourceCode, Sang-drax,
Once again you have shown clearity to everybody thanks for the input.
All you wonderful people at this forum have once again came through for me.
You all kick major @\$\$

Post # 420.
hehehehehe

Sang-drax
02-05-2005, 07:11 PM
I think this is way too tough for a college algebra courseHmm, I don't know about that...

The book says no solution.Why? It could be argued that there are no solutions if you're only working wíth real numbers, but it's a bad example if so.

To change the subject:
Here's a problem I'm currently trying to solve for a course I'm taking:

Take a function f: R->R (C1) for which
|f'(x)| <= C|f(x)| for all x€R and some positive constant C

Show that if f(0)=0, f(x)=0 for every x€R.

The special case |f'(x)|=C|f(x)| is quite easy to handle, but the general case is quite tricky. I've made some progress using integrals, but I'm having a feeling that there is an easier way.

xddxogm3
02-05-2005, 09:17 PM
Why? It could be argued that there are no solutions if you're only working wíth real numbers, but it's a bad example if so.
The book says "no solution". This includes non-real solutions also. The reason why I know that, is that the book states the answer for a problem 2 problems later on the same page is listed as having "no real solutions". It is either a typo, or a full out error.

Zach L.
02-06-2005, 01:02 AM
Depending on the scope of the book, it may equate "no solution" with "no real solution". If this is the case, it is more of a simplification, I'd say than an error.

Sang-drax, I've only spent a couple minutes looking at your problem (and its late), but my idea would be to treat it as a differential equation. We know that f(0)=0, and so, f'(0)=0, hence no slope, and any approximation of f(x) (x > 0) based on this yields f(x)=0, and then work with the limiting case as x->0. I haven't thought through it much, so I haven't considered if the absolute values will add any complication to the problem (as opposed to restricting the ranges of the functions), but it seems to me that this would be a promising direction to take. What direction have you been taking with it?

Cheers

Sang-drax
02-06-2005, 06:02 AM
I've approached it as a differential equation and found out that it works well for the special case with an equality. I'm not been able to generalize it though. I'll probably think a lot about this during the day.

Zach L.
02-06-2005, 10:26 AM
Well, you know that the maximal case is when |f'| = c|f|, and that you are dealing with strictly non-negative values, so this seems sufficient to show that the general case of |f'| <= c|f| is "bounded" between f(x) = 0 and f(x) = 0.

Sang-drax
02-06-2005, 10:57 AM
I think the inequality complicates things more than that.
Almost every function != 0 satisfies
|f'(x)|<=C|f(x)|
not that many satisfies the equality though. f(x) = Ae^Cx

Psychlow
02-08-2005, 07:20 PM
I'm the guy that xvid was helping out by posting this - thanks for assuring me that I'm not crazy, guys! :D

Personally, I'd think -4 should be listed as an answer on that equation. Last I checked, -4 was part of the real number set. :p

Sang-drax
02-09-2005, 03:57 AM
Personally, I'd think -4 should be listed as an answer on that equation. Last I checked, -4 was part of the real number set. :p
-4 is a real number, yes, but you need complex numbers to make x^(-1/2) defined for negative values of x.

It was arguments like this that introduced the complex numbers. Obviously, the first equation has a root -4, but we cannot insert x=-4 without complex numbers.

EDIT: The book is still wrong.

xddxogm3
02-09-2005, 04:49 PM
I know I said I would put the proof back in the post, but for all my searching I can not find the proof. I'm glad you all figured it out with out me putting that back in. Thanks again for everybody's input.
Psychlow - good to see you finally posting on the best forum around.
Now if you could only get away from writing in VB.
;0) hehehehe
Just Joking. VB has a purpose, I haven't found it yet, but it does.