View Full Version : Badly need some help with math
MrSparkle
2008-10-19, 12:04
I'm trying to keep up with my math class in college but its real hard because I only did up to 8th grade math. I have a test tomorrow and the teacher gave us a sample test and said if we can get all these right we'll get 100% on the real test. I could only do about 4 of the questions though. Heres the ones I don't know how to do.
Q1.) "X + 1 and X -2 are factors of X^3 + 2X^2 + AX + B
Find A and B and other factor"
So the factors of this equation are (X + 1)(X - 2)(X + C). What now though? I'm stuck.
Q2.) 2Solve
2X - 3Y = 1
X^2 - 2XY - 3Y^2 = -3"
This is just a simultaneous equation but its such a complex one I can't do it. I know I have to substitute the linear (top) equation into the quadratic (bottom) but once I've done that I'm stuck. I turn the top equation into "X = 1/2 + 3/2Y". These fractions don't make it any simpler. Heres what I get
"(1/2 - 3/2Y)^2 - 2(1/2 + 3/2Y)Y - 3Y^2 = -3" :confused::confused:
How would I simplify this? Do I have to expand the "(1/2 -3/2Y)^2" ??
Q3.) "expand (2 + square root of 3)^5 answer as A + Bsquare root of 3"
I could probably type that into a calculate couldn't I? If not I'm stuck. When you multiply a square root by itself does it become a cubic root? I'm guessing theres some sort of formula for doing this kinda question rather than just multiplying all of that by itself 5 times.
Q4.) "Find coefficient of X^10 in expansion (2X 03)^14"
I was reading up on binomial expansion and read a bit about pascals triangle but I don't know how to make a pascal triangle with this binomial.I definitely don't know how to find that coefficient in my head.
Q5.) "If roots of pX^2 - 4X + 1 are equal find P"
What are roots? Are they just the factors of the equation if it equals 0.
Q7.) "(K -2)X^2 + 2X - K = 0
prove roots are real"
Mantikore
2008-10-19, 14:44
Q1.) "X + 1 and X -2 are factors of X^3 + 2X^2 + AX + B
Find A and B and other factor"
x+1 is a factor. by the factor theorem of polynomials, if (x - n) is a factor P(n)=0. so:
P(x) = X^3 + 2X^2 + AX + B
P(-1) = 1 - a +b = 0
P(2) = 16 + 2a + b = 0
solving simultaneosly, in P(-1), re arrange so that a = 1 +b. substitute into the other equation and we get
b = -6
a = -5
Q2.) 2Solve
2X - 3Y = 1
X^2 - 2XY - 3Y^2 = -3"
This is just a simultaneous equation but its such a complex one I can't do it. I know I have to substitute the linear (top) equation into the quadratic (bottom) but once I've done that I'm stuck. I turn the top equation into "X = 1/2 + 3/2Y". These fractions don't make it any simpler. Heres what I get
"(1/2 - 3/2Y)^2 - 2(1/2 + 3/2Y)Y - 3Y^2 = -3"
How would I simplify this? Do I have to expand the "(1/2 -3/2Y)^2" ??
its the same principle, just needs a lot more work and is messy.
2X - 3Y = 1.........................(1)
X^2 - 2XY - 3Y^2 = -3.........(2)
ok, so re arrange equation 1 and we get
x = (3y+1)/2
substitute into equation 2. to answer your question, yes, you have to expand. this is to find a numerical value for y.
((3y +1)/2)^2 - 2((3y+1)/2)y - 3y^2 = -3 remember, in that fraction is a perfect square. inthe other fraction, the 2's cancel. i separated them into 3 parts, one with bold, one normal, and one italics, for ease of sight
(9y^2 + 6y +1)/4 -3y^2-y - 3y^2 = -3 now multiply everything by 4
9y^2 + 6y +1 -12y^2 - 4y - 12y^2 = -12 simplify and bring everything on one side
-15y^2 - 2y + 13 = 0 what im doing now is separating the negative side from the rest, so its easier to work with
-(15y^2 + 2y - 13) = 0 now we factorise
-((15y + 15)(15y - 13)/15) = 0
-(y + 1)(15y - 13) = 0 now solve for y
y = 1 or -13/15
now, substitute this back into equation 1 to get x
when y = 1,
x = 3/2
when y = -13/15,
x = 12/15
and thats where i got up to. i tried substituting it back into equation 2 to see which one was valid, but they both dont work. maybe a error on my part
Mantikore
2008-10-19, 15:23
FUUUUCCCKK!!! i accidentally overwrote what i had for the middle two questions
heres what i could salvage
you could multiply them out 5 times. no shame in that (hey, it works), though another method would be with binomial expansion or pascals triangle (see http://en.wikipedia.org/wiki/Pascal%27s_triangle and http://en.wikipedia.org/wiki/Binomial_expansion)
remember,
(sqrt(3))^3 = (sqrt(3))^2*sqrt(3) = 3*sqrt(3) and
(sqrt(3))^5 = (sqrt(3))^2*sqrt(3))^2*sqrt(3) = 3*3*sqrt(3) = 9*sqrt(3)
that is, if the term with sqrt(3) has an odd power, the result will have the sqrt(3) in it, if its even, they the square roots cancel each other
anyway, i got
362 + 1217*sqrt(3)
man i am royally pissed off. i took 15 minutes to write this post. anyway, ill do a short version
ok for question 4, you gotta do binomial expansion.
to do pascals triangle, the apex and numbers down the sides are all 1's. each term is then a sum of the two numbers above it.http://www.lincoln.k12.nv.us/alamo/high/Departments/Math/Pascal/ptreal1r.gif
take row 2 as an easy example. when you expand a perfect square (binomial of degree 2), youll get the coefficients 1,2,1 in the form x^2 + 2xy+ y^2
the expanded form of any binomial is a series of products of the two values in the bracket, only the first term has the first number of the same power you are trying to expand to, and the other number starting with power of 0 (meaning, its 1). with each successive term, subract 1 from the first number's power and add 1 to the second numbers power until the second number has the power you started off with and the first number has power 0
that is for (x +y)^n
a*x^(n)*y^(0) + b* x^(n-1)*y^(1) + ................. +c* x^(1)*y(n-1) + d*x^(0)*y^(n)
note a,b,c,d are the numbers in your pascals triangle whos row corresponds to the power in the triangle
again, say for a quadratic, it always has the coefficients 1 ,2, 1 as in the triangle
TL;DR (aka smart way) of dealing with your specific problem.
you know that [nCr] button on your calculator? thats the combinations button and is very useful. just punch a number, then punch [nCr], then another number.
say you have to expand some massive binomial, the terms on pascals triangle would be equal to nCr where n is the power of the binomial, and r is that number'th term.
r starts from 0, all the way up to n. notice nC0 and nCn = 1 which matches pascals triangle
eg. look at pascals triangle. check row 4. you wanna find the 3rd term. push 4C2. the answer is 6, which matches pascals triangle
in your question
to find the coefficient when power of x = 10 , find
14C10 * 2x^10*-3^(14-10)
find the coefficient, and thats your answer
actually it may be 9, not 10, i dont remember
MrSparkle
2008-10-19, 15:32
Thanks alot. I didn't know about the factor theorem.
Fuckin hell nice one. Thats the most helpful reply I've ever gotten. I asked all these same questions on a math forum and I didn't get a single helpful reply. I got a load of confusing replies that assumed I knew lots of methods and terminology that I don't know. I've been at this shit for hours trying to figure it out and I figured it out in seconds reading your post.
Mantikore
2008-10-19, 16:20
Q5.) "If roots of pX^2 - 4X + 1 are equal find P"
What are roots? Are they just the factors of the equation if it equals 0.
Q7.) "(K -2)X^2 + 2X - K = 0
prove roots are real"
these are easy if you understand just some simple theory
know the quadratic formula? you should
anyway, for a quadratic ax^2 +bx + c
x = (-b+_ sqrt(b^2 - 4ac))/2a..................(i cant do plus or minus signs :()
as you know, you cant get a square root of a negative number. you see the term inside the square root? b^2 - 4ac?
this number is very important, and is called the discriminant.
as you know, if this number is less than 0 (ie negative), there are no real roots. if you were to graph this, the parabola would not touch the x axis
if this number is 0, then you dont have to "plus or minus anything" therefore, there is only one root (that is repeated twice). if you were to draw a parabola, it would touch the x axis once.. think of graphing a perfect square
if this number is greater than 0, then we have two different roots. the graph would cut the x axis at two points
in summary
b^2 - 4ac > 0 .........two roots
b^2 - 4ac = 0........one root
b^2 - 4ac < 0 .........no roots (no real ones anyway :cool:)
pX^2 - 4X + 1
basically, equal roots means there is only one.
so
(-4)^2 - 4*P *1 = 0
16-4P = 0
P = 4 or -4
(K -2)X^2 + 2X - K = 0
basically, the disciminant >= 0 (it can have one or two roots, as long as they are real)
b^2 - 4ac >= 0
2^2 - 4*(k-2)*k >= 0
4 - 4k^2 - 8k >=0....................this is a quadratic. solve this
1 - k^2 - 2k >=0
hmm i get k = 1 +_ sqrt(8).
so im not sure how to prove it.
go easy on me, i havent done this since high school. :)
-----------------
anyway, thats all from me. if anyone wants to critique or add anything, go ahead. i lost track of time (its 3 am) and i have university tomorrow
MrSparkle
2008-10-19, 18:19
Thanks again for the help. Your a lifesaver. You probably saved me from failing my first math test. I'm gonna get one of those private tutors after this because theres so much stuff I don't know since I didn't finish high school. I have another 3 or 4 hours left to study so I should be able to retain all the methods I need to know in that space of time. I can nearly do all of these questions now.
MrSparkle
2008-10-19, 18:34
BTW I had a tab still open with your reply that you deleted. Here it is
Q3.) "expand (2 + square root of 3)^5 answer as A + Bsquare root of 3"
I could probably type that into a calculate couldn't I? If not I'm stuck. When you multiply a square root by itself does it become a cubic root? I'm guessing theres some sort of formula for doing this kinda question rather than just multiplying all of that by itself 5 times.
you could multiply them out 5 times. no shame in that (hey, it works), though another method would be with binomial expansion or pascals triangle (see http://en.wikipedia.org/wiki/Pascal%27s_triangle and http://en.wikipedia.org/wiki/Binomial_expansion)
remember,
(sqrt(3))^3 = (sqrt(3))^2*sqrt(3) = 3*sqrt(3) and
(sqrt(3))^5 = (sqrt(3))^2*sqrt(3))^2*sqrt(3) = 3*3*sqrt(3) = 9*sqrt(3)
that is, if the term with sqrt(3) has an odd power, the result will have the sqrt(3) in it, if its even, they the square roots cancel each other
anyway, i got
362 + 1217*sqrt(3)
Q4.) "Find coefficient of X^10 in expansion (2X 03)^14"
I was reading up on binomial expansion and read a bit about pascals triangle but I don't know how to make a pascal triangle with this binomial.I definitely don't know how to find that coefficient in my head.
ok. crash course in binomial theorem.
when drawing pascals triangle, start with 1 and the tip of the triangle, and the numbers running down the sides of the triangle are all 1. each number in the triangle is then equal to the sum of the numbers above it. seehttp://www.lincoln.k12.nv.us/alamo/high/Departments/Math/Pascal/ptreal1r.gif the second row has 1,2,1. theres a 2 there since the two numbers above them are 1 and 1, whose sum is 2. in row 3, the 3's are the sum of the 2 and 1 above it.
you have an error in your question. (2X 03)^14. i will assume you are talking about (2x - 3)^14.
when expanding binomials, and lets say the power (aka index) is 14. the expanded result is in the form
a*(2x)^14*3^0 + b*(2x)^13*3^1 + c*(2x)^12*3^2 + .................. + d*(2x)^2*3^12 +e*(2x)^1*3^13 + f*(2x)^0*3^14
follow the rule when expanding (m + n)^p
a*m^p*n^0 + b*m^(p-1)*n^1 +........................ + c*m^2*n^(p-1) + d*m^1*n^p
to explain this in words, the whole sequence is just the product of the two elements in the brackets, m and n. the thing is, m starts with that power p and n with the power 0 (effectively making it 1, but ill keep it there for completeness) with every step, the power of m goes down by 1 and the power of n goes up by 1 until we get m with a power of 0 (making it 1) and n with the power of p.
on top of this, we have to multiply each term with a,b,c,d.....etc. where a,b,c,d etc are the numbers in pascals triangle corresponding to the row on that triangle.
eg. if you expand a quadratic, notice how you always get x^2 + 2xy + y^2 this corresponds to row 2 on pascals triangle, with numbers 1,2,1
another note on these coefficients if you dont want to draw pascals triangle is to use the combinations button in your calculator (its the button with nCr on it) basically, you just press the power you are expanding to , then the [nCr] button, then that numbered term.
eg. to find the coefficient to multiply in the 3rd term for a binomial expansion to the power of 4, just punch
4C3, and you should get 6, which corresponds to pascals triangle
thats really how binomial theorem works
TL: DR version
HOWEVER in your case, there is a shortcut. you dont have to expand it all. you just have to find the coefficient of x^10.
using what we have learned, the term itself is just
(14C10) * 2x^10 * (-3)^(14-10)
=1001 * 162x^10
=162162x^10
so the coefficient is 162162.
be careful: the coefficient is NOT necessarily the number you get from pascals triangle. you also have to consider multiplying everything else. it sounds like common sense, but lots of people forget this.
DarthVader77
2008-10-20, 01:40
holy shit, what math class is that for?
Mantikore
2008-10-20, 03:46
holy shit, what math class is that for?
wat? i learned that stuff back in high school. :confused:
MrSparkle
2008-10-20, 14:53
Yeah its just higher level high school math. It just looks complex written on this forum because you can't type a lot of the symbols and shit. Its not simple though. I had trouble with that test. I'll find out if I passed it in a few days.
DarthVader77
2008-10-20, 20:38
oh, im in pre-calc/trig. thats probably y i dont understand it. is that calc or what?
Mantikore
2008-10-21, 03:59
oh, im in pre-calc/trig. thats probably y i dont understand it. is that calc or what?
here in australia we just called it
"maths, 2 unit and extension 1", as opposed to the other maths courses
"maths, 2 unit" (basic maths, with basic calculus and stuff)
"general maths" (shitbox nigger maths)
"maths, 2unit and extension 2" (fucking hard psycho difficult university grade maths)
MrSparkle
2008-10-21, 17:17
Here in Ireland this maths I'm doing is honours leaving cert maths I think. The leaving certs the last test you do in secondary school (high school for any American retards reading this). I didn't even do my junior cert since I left school in 3rd year (American retards: 8th grade) so theres a lot of shit I have to catch up on.
Australia must have higher education standards than America cuz my brother went to high school in America and he says you don't get maths this complicated there.
I think Ireland has one of the highest education standards in the world.
Agent 008
2008-10-21, 18:19
Take this to BLTC.
DarthVader77
2008-10-21, 21:10
Australia must have higher education standards than America cuz my brother went to high school in America and he says you don't get maths this complicated there.
I think Ireland has one of the highest education standards in the world.
americas education system is really fucked up. we probably have one of the lowest standards in the world if you dont count countries that cant even afford education
Shadout Mapes
2008-10-22, 02:42
Umm, i went to public high school in america and this looks like algebra 2 (10th grade) to me, although obviously the problems are harder since its a college course. It's also more specifically oriented towards algebra, whereas the classes i took were more on functions/roots (i think we covered binomial expansions for only like a week or so). i think the main difference is just that american high school maths are geared towards prepping you specifically for calculus. that being said, america is still behind most other countries, which cover much more material in high school/undergraduate college.
Mantikore
2008-10-22, 03:18
americas education system is really fucked up. we probably have one of the lowest standards in the world if you dont count countries that cant even afford education
hmm ive found although american schooling is kind of bad, their college/university grade maths is really advanced
MrSparkle
2008-10-22, 16:59
Yea I think America makes up for their crap high schools with their universities.
DarthVader77
2008-10-22, 21:11
Yea I think America makes up for their crap high schools with their universities.
definitely