I printed out the 2003 tests from this site, and i've got quite a few questions, many concerning roots. Sorry if the questions are kind-of repetitive.
1.) The largest root of (6x^2) - x - 15 = 0 (answer is 5/3)
2.) If (x^2)+2xy+(y^2) = 16 and x = 6, then the largest value of y = ? (answer is -2)
3.) If x = 3 and y = 5 then (x-y)(x^2+xy+y^2) =? (-98)
-i don't really need a trick for that one, but i see that equation quite a bit, so i thought there might be one.
4.) The smallest root of (x^2)-13x-48=0 is ? (-3)
5.)If r1 and r2 are roots of 3x^2 - 5x = 4, then r1 * r2 =? (-4/3)
6.) If equation (2x^3) - (bx^2) + cx=d has roots r, s, and t. If rst = 3.5, then d =? (7)
7.) If r, s, and t are the roots of the equations
(2x^3)-(4x^2)+6x = 8, then rs+rt+st =? (3)
-sorry about it being so long and confusing, but there's a lot of problems dealing with roots that i have no clue how to do. Thanks for everything guys,
-Cole
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1.) You just need to factor the part of the equation set to zero. We find factors of ac=6*-15=-90 adding to b=-1. Now, it's pretty clear from that that we need to get -10 and 9, just from intuition. So we set it up so that the 3's in a and c will be multiplied, forming 9, and the 2 and -5 in a and c will multiply to get our -10. 6x^2-x-15 factors to (3x-5)(2x+3). If (3x-5)(2x+3)=0, then there are two possibilities. First, 3x-5=0 and x=5/3, and second, 2x+3=0 and x=-3/2. Take the larger.
2.) This is a simple factoring pattern. You could take the other approach, but you'll find you end up doing the same thing. (x^2)+2xy+(y^2)=(x+y)(x+y)=(x+y)^2. Take the square root, because you're basically given (x+y)^2=16, and you get |x+y|=4. Split it into two equations, and you get x+y=4 or x+y=-4. Substitute 6 in for x (the other piece of information you're given) and you have either y=-10 or y=-2. Take the larger.
3.) Again, this is a factoring pattern, the difference of cubes. (x-y)(x^2+xy+y^2)=x^3-y^3. substitute values and you get 27-125, or -98.
There is a complimentary factoring pattern, the sum of cubes. It is: a^3+b^3=(a+b)(a^2-ab+b^2).
4.) This is like question 1, but easier to think about because here a=1. Find factors of ac=1*-48=-48 adding to b=-13. We get 3 and -16. Set it up with 1 and 3 multiplied to 3, and 1 and -16 multiplied to -16 and we get (x^2)-13x-48=(x+3)(x-16)=0, for which one factor must be zero. Set each equal to zero, and your roots are x=-3 and x=16. Take the smaller.
5.) You could factor this. DON'T. For quadratics, the product of roots=c/a. Subtract 4 to get it in standard from, put your c over your a, and you have the answer.
Sorry, I forgot how to do the other stuff... I'll look it up later if nobody else comes.
6) Product of the roots of an odd degree polynomial (starting with x^3, x^5, etc.) is (-constant term/first term). Therefore in ax^3 + bx^2 + cx + d = 0, product of the roots is -d/a. They tell you that the product of the roots is 3.5, so -d/2 = 3.5. d must equal -7. Since it is on the opposite side of the equation, switch signs to get d=7.
7)This is sum of the roots taken two at a time, which is c/a giving 6/2 giving 3.
If you wish to dive deeper into the reletionships of coefficients and roots of a polynomial, read Aaron Goldsmith's posts in the texasmath forum about #39 on Math district 2.
I do practice quite a bit, but i've only got a limited number of tests. I am going faster, but i think its because i've taken the same tests so many times and i'm remembering the answers.
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Don't take me seriously or i'll hunt you down and kill you.
Yeah, But thats not a good habbit to get into. You need to be ready for anything, so you should broaden your horizon with any test you can get your hands on. Make sure you understand why you miss what you miss, and pretty soon you wont be missing those problems. Then, suddenly you begin to see patterns in the test, and you go faster and faster.... Suddenly, youre scoring up there with Vinay, Quincy, (not sam) and R. I plan on working more for next year than this. The most I can hope for in our region (We got Medina Valley to deal with) is a 4th place. But hey, I'll take it. If NumSense is your only event right now, PRACTICE. There is no other way to improve your score enough to be competitive.
Aye, PRACTICE!!! And also, TRY NOT TO LOOK FOR SHORTCUTS ELSEWHERE, CHALLENGE YOURSELF TO FIND THEM! ONCE YOU SUCCEEDED, THEY SINK IN AND STICK WITH YOU. You can find similar questions among those packets, compare their answer, and use your NUMBER SENSE to make them out, if you don't have the leisure to prove them. That's how I started off, teaching my coach those shortcuts.
I was taking a practice regional test, and i realized that i forgot how to do the factorial problems. I left my test at home, (i'm at my grandma's) so i don't have any examples, but most of you probably know what i'm talking about. Also, i'd like to know if they appear in the same pattern every time. Thanks everybody.
-Cole-
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Don't take me seriously or i'll hunt you down and kill you.
wait... wouldnt it be like (8*7*6*5*4*3*2*1/7*6*5*4*3*2*1) + (6*5*4*3*2*1/7*6*5*4*3*2*1) which equals 8 + 1/7. so your answer should be 8 (1/7). Right?