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InvisibleCapers
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How to tell the number of solutions to an expression with negative exponents?
    #15924344 - 03/09/12 12:23 PM (12 years, 21 days ago)

Ok, I need a little help with some general math concepts.

If I understand correctly a polynomial is an expression with variables and exponents, but the exponents have to be positive whole numbers and there can be no variables in the denominator of a term.

To my understanding, you can look at a polynomial equation and tell how many solutions it has based on the highest exponent. For example:

x^3 + x^2 + x + 1 = 0 would be a polynomial with three solutions.

Am I correct in this understanding?


My next question is about equations that aren't polynomials. How do you tell by looking at them how many solutions there are?

For example:

-3.4x^-3 + .25 = 0

I already found one solution to be x = (.25/3.4)^(-1/3), which is approximately x = 2.39. However, I suspect there are more answers.

Could someone explain this to me please?

Edited by Capers (03/09/12 12:26 PM)

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InvisibleDieCommie

Registered: 12/11/03
Posts: 29,258
Re: How to tell the number of solutions to an expression with negative exponents? [Re: Capers]
    #15924533 - 03/09/12 01:01 PM (12 years, 21 days ago)

Hint:  A negative exponent on one side of the equation is the same as a positive exponent on the other side.  Note that generally if you have negative and positive exponents on the same side that do not cancel you have a very hard equation to solve by hand.  Most likely all your exponents can be moved to the same side and solved with your known techniques.  Also note that even roots have two solutions while odd roots have one solution.

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InvisibleCapers
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: DieCommie]
    #15924686 - 03/09/12 01:35 PM (12 years, 21 days ago)

Wait, how is a negative exponent on one side of the equation the same as a positive exponent on the other side?

Using my example:

-3.4x^-3 + .25 = 0
x^-3 = (.25/3.4)
x = (.25/3.4)^(-1/3)

right? How would the -3 exponent on the left equal a positive exponent on the right?

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InvisibleDieCommie

Registered: 12/11/03
Posts: 29,258
Re: How to tell the number of solutions to an expression with negative exponents? [Re: Capers]
    #15924702 - 03/09/12 01:38 PM (12 years, 21 days ago)

At this point,

Formula: 0

You can multiply both sides by x3 and get rid of all negative exponents.

Make sure you understand the arithmetic behind exponents, your last line doesn't look right.

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Offlinecircastes
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: DieCommie]
    #15924790 - 03/09/12 02:01 PM (12 years, 21 days ago)

Think about the shape of the graphed equation. How many times does it touch the x-axis (such that y=0)? (ax^2 + bx + c) is a parabola, which can have 1,2 or no solution. 1 solution occurs if it is touching the x-axis, 2 solutions occur if it is below the x-axis as the two arms reach past the x-axis, no solution occurs if it is above the x-axis.

Just sketch it with a calculator or use logic to quickly reason if it will be touching the x-axis / below it / above it. Voila, you know how many answers there are.


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InvisibleCapers
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: DieCommie]
    #15924793 - 03/09/12 02:01 PM (12 years, 21 days ago)

Ok, I see. Thanks.

Not that it makes a difference in my understanding of your point, but I'm pretty sure the arithmetic was correct.

x^3 = y is the same as x = y ^(1/3), right?

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Offlinecircastes
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: circastes]
    #15924796 - 03/09/12 02:02 PM (12 years, 21 days ago)

btw -3.4x^-3 + .25 = 0

is

-3.4x^-3 + 0x^2 + 0x + .25 = 0

so is still a polynomial...


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TESTED
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InvisibleCapers
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: circastes]
    #15924799 - 03/09/12 02:03 PM (12 years, 21 days ago)

Quote:

circastes said:
Think about the shape of the graphed equation. How many times does it touch the x-axis (such that y=0)? (ax^2 + bx + c) is a parabola, which can have 1,2 or no solution. 1 solution occurs if it is touching the x-axis, 2 solutions occur if it is below the x-axis as the two arms reach past the x-axis, no solution occurs if it is above the x-axis.

Just sketch it with a calculator or use logic to quickly reason if it will be touching the x-axis / below it / above it. Voila, you know how many answers there are.




Yah thanks, I do understand that that's an option, but I wanted to know how to tell just by looking at the equation. I know it's rare, but you could have a tricky graph that has more roots beyond your present viewing window, which could trick you into thinking there are less answers than in reality.

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Offlinecircastes
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: circastes]
    #15924802 - 03/09/12 02:04 PM (12 years, 21 days ago)

Look up "index laws" :thumbup:


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InvisibleCapers
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: circastes]
    #15924805 - 03/09/12 02:04 PM (12 years, 21 days ago)

Quote:

circastes said:
btw -3.4x^-3 + .25 = 0

is

-3.4x^-3 + 0x^2 + 0x + .25 = 0

so is still a polynomial...




Oh ok. I guess that makes sense now that I know negative exponents on one side are positive exponents on the other side.

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InvisibleCapers
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: circastes]
    #15924810 - 03/09/12 02:06 PM (12 years, 21 days ago)

Quote:

circastes said:
Look up "index laws" :thumbup:




Ok, I'm about to catch my favorite radio news show. I'll try to look it up afterward. Could you please give me a brief explanation of what index laws are about or why it would be useful to me?

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Offlinecircastes
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Re: How to tell the number of solutions to an expression with negative exponents? [Re: Capers]
    #15924834 - 03/09/12 02:12 PM (12 years, 21 days ago)

They are concerned with exponents.

ie. x^2 * x^2 ---> add the exponents = x^(2 + 2) = x^4

ie. x^0 = 1

If you want to take an exponent out of an equation by multi/div both sides, think of the index law(s) that will do it.

This:

-3.4x^-3 + .25 = 0
x^-3 = (.25/3.4)
x = (.25/3.4)^(-1/3)

no...

-3.4x^-3 + .25 = 0
x^-3 = (.25/3.4)
1 = (.25/3.4)*(x^3)    <---
1/(.25/3.4) = x^3
then cube root

I think. I can't tell if that's the same.

Look up index laws anyway. Will definitely help!


--------------------
My solitude...
My shield...
My armour...

TESTED
WITH
FULL
FORCE

Edited by circastes (03/09/12 02:18 PM)

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