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Math Trouble


MarlinFan10

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Let's say we lost a packet containing information on polynomials and we have a question on a practice test involving finding the sum of the squares of the roots of a 10th order polynomial. What should one do in that situation?

 

Yea its kinda almost too simple to be on a competition test, but it works when people like me forget dumb things like this.

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Problem in question:

 

What is the sum of the squares of the zeroes of:

 

12x^10 + 4x^9 - 3X^8 - 5?

 

a. 1/3

b. 7/18

c. 11/18

d. 1/9

e. NOTA

 

(x^10 means x to the 10th power)

648618[/snapback]

hmm....the sum of the squares of the zeroes.

 

Let me try this:

 

12x^(10)+4x^(9)-3x^(8)-5

 

OK, first I need to find the zeroes. (I am using a calculator like you are allowed to in the competitions from what I understand.)

 

This is an even polynomial so there are likely and even amount of zeroes.

 

x= -1, approx. .9105

 

I got approx. 1.829 which would be e. NOTA

 

However, I don't feel this is right.

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u gotta find the logs and stuff. turn it into ln and log or someshit. i dont want to think too much but i think its somehting like this for ex.

 

10log12 + 9log4

-------------------

8log3 - log5

 

the --------- is a division symbol.

 

then u plug it in a graphing calc (i lost mines) and it should give u the answer. its something like that or with ln . i forgot. hope that little bit helps dood.

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I think I am missing around 8 other zeroes :lol

648802[/snapback]

Finding the zeroes would cause death during a competition. Getting the sum of the zeroes and the sum taken two at a time and then using the formula is the only way to go.

 

Anyways, the sum of the zeroes is -1/3. Sum of zeroes is -B/A

 

The sum taken two at a time is -1/4. Sum taken 2 at a time is found by going 2 terms to the right of A and putting that coefficient over A.

 

Therefore the answer is :

(-1/3)(-1/3)-2(-1/4)

1/9+1/2

 

11/18, C. I hope i did that right. :thumbup

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New problem, and its probability(my weak point most of the time). I was shown how to do this type a year ago but always get these things wrong:

 

John, George, and Paul are playing a game of cards with the preceding order. John has a probability of winning of 1/5 each turn, George of 1/4, and Paul of 2/9. If they play until someone wins, then what is the probability that John will win the game?

 

a. 1/5

b. 3/8

c. 7/75

d. 3/32

e. NOTA

 

:blink:

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I think I am missing around 8 other zeroes :lol

648802[/snapback]

Finding the zeroes would cause death during a competition. Getting the sum of the zeroes and the sum taken two at a time and then using the formula is the only way to go.

 

Anyways, the sum of the zeroes is -1/3. Sum of zeroes is -B/A

 

The sum taken two at a time is -1/4. Sum taken 2 at a time is found by going 2 terms to the right of A and putting that coefficient over A.

 

Therefore the answer is :

(-1/3)(-1/3)-2(-1/4)

1/9+1/2

 

11/18, C. I hope i did that right. :thumbup

648931[/snapback]

 

 

well when u get older and can use a graphing calculator the way i did it is the best and fastest way to do it.

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I think I am missing around 8 other zeroes :lol

648802[/snapback]

Finding the zeroes would cause death during a competition. Getting the sum of the zeroes and the sum taken two at a time and then using the formula is the only way to go.

 

Anyways, the sum of the zeroes is -1/3. Sum of zeroes is -B/A

 

The sum taken two at a time is -1/4. Sum taken 2 at a time is found by going 2 terms to the right of A and putting that coefficient over A.

 

Therefore the answer is :

(-1/3)(-1/3)-2(-1/4)

1/9+1/2

 

11/18, C. I hope i did that right. :thumbup

648931[/snapback]

 

 

well when u get older and can use a graphing calculator the way i did it is the best and fastest way to do it.

649057[/snapback]

Graphing calculaters are not allowed in competition. :whistle

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I think I am missing around 8 other zeroes :lol

648802[/snapback]

Finding the zeroes would cause death during a competition. Getting the sum of the zeroes and the sum taken two at a time and then using the formula is the only way to go.

 

Anyways, the sum of the zeroes is -1/3. Sum of zeroes is -B/A

 

The sum taken two at a time is -1/4. Sum taken 2 at a time is found by going 2 terms to the right of A and putting that coefficient over A.

 

Therefore the answer is :

(-1/3)(-1/3)-2(-1/4)

1/9+1/2

 

11/18, C. I hope i did that right. :thumbup

648931[/snapback]

 

 

well when u get older and can use a graphing calculator the way i did it is the best and fastest way to do it.

649057[/snapback]

Graphing calculaters are not allowed in competition. :whistle

649093[/snapback]

 

 

i just noticed that scientific calculators have logs and ln too. if thats allowed in the compitition u should do it that way. takes 2 min

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and knowing this will help you do what in real life?

648779[/snapback]

 

Absolutely nothing.. In my opinion, it actually works against you to know all this meaningless crap.

 

Crap like this is taking up valuable space in your head that you could use for other things. Unless you plan on being a math teacher, this is horse-shite. I'm a Systems Analyst and I wish I could pay someone to dial into my head and remove all the excess crap that I've learned throughout the years, in hopes of clearing up some space for more technology related information... Oh well....

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