January 6, 200521 yr 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.
January 6, 200521 yr Author 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)
January 6, 200521 yr Author I'm dying. Someone on this site has to know how to find the sum of the squares of the roots of a polynomial. :unsure
January 6, 200521 yr Author First of all I would like to thank you all for being useless. :plain The sum of the squares of the roots is: The sum of the roots squared minus twice the sum of the roots taken two at a time. The mystery is solved, and the world is saved from destruction! :elefant
January 6, 200521 yr and knowing this will help you do what in real life? not a knock on you at all, just curious.
January 6, 200521 yr Author and knowing this will help you do what in real life? not a knock on you at all, just curious. 648779[/snapback] Do better in math competition this year of course. :whistle
January 6, 200521 yr 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.
January 6, 200521 yr Someone graph it with a calculator, find all the zeros, write down where they occur, and add them up.
January 6, 200521 yr 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.
January 6, 200521 yr Author 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
January 6, 200521 yr Author 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:
January 6, 200521 yr Thoughts on the second one: 1) Where's Ringo? 2) The question is ambigiuous. Does "each turn" imply the entire game? If not, then you don't know the criteria for winning the overall game.
January 6, 200521 yr 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.
January 6, 200521 yr Author 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
January 6, 200521 yr Gosh I hate math..... *wonders why the hell she needs to take math if she's studying english education*
January 7, 200521 yr Author Well it took me a while but I figured this out. You set up an infinite geometric series with first term 1/5 and ratio 7/15. The final answer is 3/8.
January 7, 200521 yr 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
January 7, 200521 yr Author Bleh I did about 15 of these last years, but I couldn't remember how to do these when I came across it. How many zeros are at the end of 1994! (1994 factorial).? Also: What is the remained when 7^99 is divided by 13. I need to find out how to do these two things and I'll be set.
January 7, 200521 yr 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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