QUESTIONS- HOPE - INEQUALITIES,RATIO-PROPORTION

QUESTIONS - ANSWERS - HOPE - INEQUALITIES-RATIO & PROPORTION

Hope has left a new comment on your post "QUESTIONS - COMPLEX ANALYSIS":

Solve the following polynomial inequality and give the answer in interval notation. 2x^3 + 4.5x^2 < 0
X^2(2X+4.5)<0
X^2 BEING A PERFECT SQUARE IS ALWAYS >=0.
HENCE
2X+4.5 <0
2X<-4.5
X<-4.5/2 = - 2.25…..

HENCE IN INTERVAL NOTATION
X SHOULD LIE IN (-INFINITY, - 2.25)



What kind of variation is demonstrated by the following function? f(x) = (x / 100) + 1
STRICTLY SPEAKING F(X) DOES NOT FALL UNDER CATEGORY OF “ VARIATION BY DIRECT OR INDIRECT PROPORTION”, BUT
F(X)-1 VARIES DIRECTLY AS X WITH A CONSTANT OF PROPORTION EQUAL TO 1/100.THIS IS BECAUSE ..
X=………………......…100…………… 200……………300….ETC
F(X)-1=(X/100)……1………………..2………………3…ETC.THAT IS
[F(X)-1]/X = CONSTANT ALWAYS.(=1/100 HERE) SO WE SAY
F(X)-1 VARIES DIRECTLY AS X .
BUT NOTE THAT IF WE TAKE ONLY F(X)
X=…………………………......100……………200………………300
F(X) =(X/100)+1…………2…………………3……………….4…ETC
AND F(X)/X = 2/100,3/100,4/100 ETC..THAT IS VARIABLE..NOT CONSTANT.
BUT IS YOU ARE TAUGHT TO CALL 1/100 AS CONSTANT OF PROPORTIONALITY ..THEN YOU MAY CALL SO.



For the constant of proportionality in the following function I get 100. Is that correct?
f(x) = 100x + 6


TO TELL IN ONE WORD 'CORRECT'. BUT TO EXPLAIN ...

WELL YOU CAN SAY SO,IF THAT IS THE PRACTICE THERE. NORMALLY IT IS CALLED SLOPE.
WHY I DO NOT ADVOCATE CALLING 'CONSTANT OF PROPORTIONALITY' IS THAT ,
SOME STUDENTS MAY MISINTERPRET IT AS TO CONCLUDE THAT Y [OR F(X) AS YOU PUT
IT] IS PROPORTIONAL TO X AND VARIES AS X WHICH IS NOT CORRECT.
ACTUALLY Y-6 IS PROPORTIONAL TO X AND VARIES AS X.
IF Y IS TO BE REALLY PROPORTIONAL TO X THEN THE EQN. HAS TO BE
Y=100X.......ONLY


In the following function, y varies as what power of x? y = 5 / x³ Thank you for all the help you have given me.
SINCE Y = 5/X^3 = CONSTANT/X^3…..WE SAY
Y VARIES INVERSELY AS X^3…THAT IS TO THE THIRD POWER OF X ……………….OR…………………
Y VARIES DIRECTLY AS X^(-3) …THAT IS TO THE –3 POWER OF X .

QUESTIONS - ANSWERS - EQUATIONS

QUESTIONS - ANSWERS - EQUATIONS
Find an equation of the form y = ax^2 + bx + c whose graph passes through the points (1,-2) (2,-1), and (3,4).
SUBSTITUTING THE VALUE OF X FROM POINTS(X,Y) IN THE GIVEN EQN. FOR Y WE GET
A+B+C=-2...................1
4A+2B+C=-1.......................2
9A+3B+C=4............3
EQN.2-EQN.1
3A+B=1................4
EQN.3-EQN.2
5A+B=5.................5
EQN.5 - EQN.4
2A=4
A=2
SUBSTITUTING IN EQN.4....
6+B=1
B=-5
SUBSTITUTING IN EQN.1
2-5+C=-2
C=1
HENCE THE EQN.IS
Y = 2X^2-5X+1

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY
Let P1 be the plane defined by the points:
Q=(-1,-2,-3), R=(-3,1,-5) and S=(2,-5,-4)
Find the equation of the plane perpendicular to P1 and containing the line:
L1 SAY =(x,y,z) = (2,-2,-3) + t (2,-3,0)
EQN. OF REQUIRED PLANE IS
A(X-2)+B(Y+2)+C(Z+3)=0..............1..WITH DRS OF A,B,C.
DRS OF L1 ARE 2,-3,0...IT IS PERPENDICULAR TO NORMAL O REQD.PLANE.HENCE
2A-3B=0.......A=1.5B.........2
DRS OF NORMAL TO PLANE P1 IS GIVEN BY CROSS PRODUCT OF JOINS OF QR AND RS
DRS OF QR = [-3+1,1+2,-5+3]= [-2,3,-2]
DRS OF RS = [2+3,-5-1,-4+5]= [5,-6,1]
QR X RS = [-2i+3j-2k]X[5i-6j+k]= 12k+2j-15k+3i-10j-12i=-9i-8j-3k
THIS SHOULD BE PERPENDICULAR TO NORMAL OF REQD.PLAN.HENCE
-9A-8B-3C=0......9A+8B+3C=0...........3
9*1.5B+8B+3C=0
21.5B=-3C
C=-21.5B/3
TAKING B AS 6 , WE GET A = 9...C=-43
HENCE EQN.OF REQD PLANE IS
9(X-2)+6(Y+2)-43(Z+3)=0
9X+6Y-43Z-135=0

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY
Find the equation of the plane=P SAY, containing the line L1, and parallel to the line L2, where:
L1: (x,y,z) = (-2,-1,1)(Q SAY) + t (-3,0,2).
L2: is the intersection of the planes 2x-2y-3z=-5 and 1x-3y-3z=-3
Write your answer in the form -4x+By+Cz=D, and give the values of B, C and D as your answer.
EQUATION OF P THROUGH Q IS
A(X+2)+B(Y+1)+C(Z-1)=0..............................1
DRS OF NORMAL ARE A,B,C.........NORMAL PERPENDICULAR TO L1
-3A+2C=0........A=2C/3 ...............2
IF DRS OF L2 ARE L,M,N THEN
2L-2M-3N=0...............3
L-3M-3N=0.................4
|-2,-3|
|-3,-3|=6-9=-3
|-3,2|
|-3,1|=3
|2,-2|
|1,-3|=-4
L:M:N=-3:3:-4
HENCE
-3A+3B-4C=0...................5
-3*2C/3 +3B-4C=0
3B=6C
B=2C
A=2C/3.....OR.....TAKING C = 6....A=4,B=12
EQN OF PLANE IS
4(X+2)+12(Y+1)+6(Z-1)=0
-4X-12Y-6Z=14

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY
Find the shortest distance between the two lines:
L1 SAY..(x,y,z) = P SAY(2,-1,2) + t(-3,-2,-2)
L2 SAY (x,y,z) = Q SAY(-3,4,1) + s(0,2,0)
IT IS PROJECTION OF PQ ON DIRECTION(L,M,N SAY) PERPENDICULAR TO L1 AND L2
-3L-2M-2N=0........1
2M=0
3L=-2N
N=-1.5L
LET L=2...M=0....N=-3
PQ IS (-3-2,4+1,1-2)=(-5,5,-1)
S.D.= PROJECTION OF PQ ON (2,0,-3) DIRECTION
[-5*2+1*3)/SQRT(2^2+3^2)=-7/SQRT(13)

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY
Find the point P on the plane 0x+1y-3z=8 closest to the point Q(2,-3,3).
P = ( , , )?????(p,q,r) say
it on plane.hence
q-3r=8...............1
PQ is perpenicular to plane...or parallel to its normal
drs of plane = (0,1,-3)
drs of PQ = [p-2,q+3,r-3]
(p-2)/0 = (q+3)/1 = (r-3)/-3
p=2
r-3 = -3(q+3)=-3q-9
3q+r=-6..............2
3*eqn.1-en.2
-10r=30
r=-3
3q=-3
q=-1
hence p is (2,-1,-3)

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY

QUESTIONS - ANSWERS - VECTORS - 3D GEOMETRY
Find the shortest distance between the plane(P SAY) 2x+4y-2z=6
and the line(L SAY) (x,y,z) = (Q SAY)(-1,-2,4) + t(-2,-1,-4).
Distance =????
IF L IS NOT PARALLEL TO P THEN S.D=0 AS IT INTERSECTS THE PLANE.
IF L||P NORMAL TO PLANE IS PERPENDICULAR TO LINE
DRS OF NORMAL TO PLANE ...(2,4,-2)
DRS OF LINE ....(-2,-1,-4)
2*-2+4*-1+-2*-4=-4-4+8=0...HENCE L||P
SO TAKE ANY POINT ON LINE AND FIND ITS
PERPENDICULAR DISTANCE FROM THE PLANE.
ANY POINT ON LINE IS (-1,-2,-4)
S.D = [2*-1+4*-2+(-2)(-4)-6]/SQRT(2^2+4^2+2^2)
= -8/SQRT(24)=-2SQRT(6)/3

QUESTIONS - FUNCTIONS

QUESTIONS - ANSWERS - FUNCTIONS
Anonymous has left a new comment on your post "QUESTIONS - FUNCTIONS":

Determine the horizontal asymptote of the following rational function? f(x) = 4x / (x2 – 3)
HORIZONTAL ASYMPTOTE IS OBTAINED BY FINDING LIMIT OF Y AS X TENDS TO INFINITY.
HERE WE HAVE LIMIT AS ZERO SINCE
4X/(X^2-3) = 4/[X-(3/X)]...WHICH TENDS TO ZERO AS X TENDS TO INFINITY.
HENCE Y=0 THAT IS THE X AXIS IS A HORIZONTAL ASYMPTOTE.

what kind of asymptote occurs for the following rational function? f(x) = (10x³ + x²) / [ 4(x² - 1) ]

1.VERTICAL ASYMPTOTES:

Y TENDS TO INFINITY AS X TENDS TO + OR -1 .
HENCE
X=1 AND X=-1 ARE THE 2 VERTICAL ASYMPTOTES.

2.HORIZONTAL ASYMPTOTE:

AS X TENDS TO INFINITY Y TENDS TO INFINITY.HENCE THERE IS NO H.A.

3.INCLINED ASYMPTOTE:-
A.RIGHT LINE
Y/X = (10X^3+X^2)/(4X^3-X)=[10+(1/X)]/[4-(1/X^2)] TENDS TO 10/4 = 2.5 = K SAY
AS X TENDS TO +INFINITY.
Y-KX = [(10X^2+X)/(4X^2-1)]-2.5X = [10X^2+X-2.5X(4X^2-1)]/[4X^2-1]
=(3.5X)/(4X^2-1)...TENDS TO ZERO = L SAY.. AS X TEND TO +INFINITY.
HENCE THERE IS AN INCLINED ASYMPTOTE GIVEN BY
Y=KX+L=2.5X+0=2.5X
B.LEFT LINE.
WE FIND SAME LIMITS AS ABOVE FOR X TENDING TO -INFINITY.HENCE THERE
IS NO LEFT INCLINED ASYMPTOTE.

QUESTIONS - FUNCTIONS

QUESTIONS - ANSWERS - FUNCTIONS
On 11/28/06, Anonymous wrote:
> Anonymous has left a new comment on your post "QUESTIONS - POLYNOMIALS":
>
> If you have the time to lend some help, I'd be grateful. 1) Determine the
> domain of the following rational function. Give the answer in set notation
> and in interval notation. f(x) = 15 / (x + 10)
DOMAIN ALL THE VALUES, THE INDEPENDENT VARIABLE X CAN TAKE , WITHOUT
MAKING THE FUNCTION MEANING LESS.
DENOMINATOR X+10 SHOULD NOT BE ZERO AS IT WILL MAKE THE FUNCTION
MEANING LESS.
HENCE DOMAIN IS ALL REAL VALUES OF X EXCEPT X=-10
IN SET NOTATION
(- INFINITY,-10)U(-10,+INFINITY)


2) What are the vertical
> asymptotes of the following rational function? f(x) = (x+3) / (x2 + x –20) = Y SAY
VERTICAL ASYMPTOTES ARE THOSE VALUES OF X WHICH MAKE THE FUNCTION TEND
TO INFINITY.
Y = (X+3)/(X^2+5X-4X-20)=(X+3)/[X(X+5)-4(X+5)] = (X+3)/[(X+5)(X-4)]
HENCE AS X TENDS 4 AND -5 ,Y TENDS TO INFINITY.
X-4 =0 ...OR....X=4...AND
X+5=0...OR.....X=-5 ARE THE 2 VERTICAL ASYMPTOTES

QUESTIONS - SHERILL - DIFFERENTIALS

QUESTIONS - ANSWERS - SHERILL - DIFFERENTIALS

sherill has left a new comment on your post "QUESTIONS - ANGELA - QUADRATIC EQUATIONS":

A state game commission introduces 50 deer into newly acquired state lands. The population N of the herd can be modeled by N=10(5+3t)/1+0.04t where t is time in years. Use differentials to approximate the change in the herd size from t=5 to t=6 Please help
PLEASE CLARIFY
1.THE GIVEN EQN. IS N=10[(5+3T)/(1+0.04T)]..AND YOU WANT INTEGRATION TECHNIQUES OR DIFFERENTIALS.GIVE YOUR MAIL ADDRESS TOO.

QUESTIONS - CDOG - LINEAR ALGEBRA

QUESTIONS - ANSWERS -CDOG -LINEAR ALGEBRA

On 11/19/06, cdog wrote:
> cdog has left a new comment on your post "QUESTIONS - MAXIMA MINIMA":
>
> QUESTION ABOUT LINEAR DEPENDENCE AND LINEAR INDEPENDENCE. Can someone please
> explain to me (like I am a 5-year old) the basic difference between linear
> independence vs. linear dependence. Perhaps, I'm reading my Linear Algebra
> text incorrectly, but they appear to be a contradiction in terms. A set is
> dependent if the determinent of the coefficient matrix=0? Otherwise, the set
> is independent? Is this a correct statement?
WELL AS DESIRED BY YOU, THIS ASPECT STARTING WITH SIMPLE EXAMPLES WAS POSTED UNDER MATHS-6.EQUATIONS IN OCT '06,WHICH IS REPRODUCED BELOW.
QUOTE:
Simultaneous equations involving 2 unknowns - x and y are often referred by school students for different methods of solution and more importantly to find whether they are consistent, independent or inconsistent. Let us see 3 simple examples and elaborate on this particular aspect of existence and uniqueness of solution.

1. x+y =2 and x-y =0
2. x+y =2 and 2x+2y = 4
3. x+y =2 and 2x+2y =3

We adopt 2 methods to determine this aspect.

1.Algebraic method

2.Graphic method


I.Independent equations

Test 1 . Algebraic method.

Find the ratios of coefficients of unknowns in the 2 given equations

In ex.1…it is

1/1 =1 for x

and

1/-1 =-1 for y.

the ratios are not equal

IF THE RATIOS ARE DIFFERENT,THEY ARE INDEPENDENT EQUATIONS AND WE GET UNIQUE SOLUTION.

Here we find that x = y = 1 is the one and only one solution.

In ex.2….it is

1 / 2 for x

and

1 / 2 for y

the ratios are equal.

IF THE RATIOS ARE EQUAL THEN THEY ARE DEPENDENT EQUATIONS AND WE CAN HAVE TWO POSSIBILITIES.

A. We may have infinite solutions

We test for this possibility by finding the ratio of constant terms. Here the ratio is 2 / 4 = 1 / 2 which is same as the ratio of coefficients of unknowns.

IF RATIO OF COEFFICIENTS OF UNKNOWNS IS SAME AS THE RATIO OF CONSTANT TERMS THEN THEY ARE DEPENDENT / CONSISTENT AND WE HAVE INFINITE SOLUTIONS.

Here we have x = y = 1 or x = 0 and y =2 or x=2 and y=0 etc..are all possible solutions. In-fact any set of values x = x and y = 2 – x is a solution.

PLEASE NOTE THE COMMENTS UNDER EQUATIONS AND IDENTITIES IN THIS CONTEXT

B. We may have no solution.

IF RATIO OF COEFFICIENTS OF UNKNOWNS IS NOT SAME AS THE RATIO OF CONSTANT TERMS THEN THEY ARE CALLED INCONSISTENT AND WE HAVE NO SOLUTION.

In ex.3..it is

1 / 2 for x

1 / 2 for y

and

2 / 3 for constants which is not equal to the above ratio.

Hence the equations are in-consistent and we have no solution.

Test 2…Graphic method.

Draw graphs of the 2 equations on the same scale on the same paper.

If the 2 graphs intersect , then the point of intersection is the solution.

In Ex.1 we find that the 2 graphs…straight lines in fact intersect
at x = y = 1.Hence they are independent equations and the point of intersection viz.x = y = 1 is the solution.

In Ex.2 …we find that the 2 graphs ….straight lines in fact are coincident. Hence they are dependent/consistent and any point on the line is a solution. Say x = y = 1 or…x=2 and y =0 etc..

In Ex.3...we find that the 2 graphs...straight lines in-fact are parallel and do not intersect.hence, they are inconsistent equations and there is no solution.
UNQUOTE:
NOW COMING TO YOUR POINT OF DETERMINANT OF COEFFICIENT MATRIX BEING ZERO OR MORE PRECISELY AS YOU WOULD KNOW LATER,ITS RANK BEING LESS THAN ITS ORDER.
IN THE ABOVE EXAMPLES,TO EXPLAIN TO SCHOOL STUDENTS,WE TERMED IT AS RATIO OF COEFFICIENTS.DETERMINANTS OF THE 3 COEFFICIENT MATRICES IN THE ABOVE EXAMPLES ARE,
EX.1
|1, 1|
|1,-1| = 1*(-1)-(1*1)=-2
EX.2
|1,1|
|2,2| = 1*2 - 1*2 =0
EX.3
|1,1|
|2,2| = 1*2 - 1*2 =0

SO THE RULE IS CORRECT
IF THE DETERMINANT OF COEFFICIENT MATRIX IS ZERO,THEN THE EQNS. ARE TERMED DEPENDENT
BUT WHETHER THEY ARE CONSISTENT OR NOT DEPENDS ON THE CONSTANT TERMS AS BROUGHT OUT BY ME IN THE ABOVE LESSON. THE TRANSLATION OF THIS TO MATRICES,YOU WILL LEARN SOON..
A FURTHER GENERAL DEFINITION FOR MORE THAN 2 EQUATIONS CAN BE AS FOLLOWS.
IF WE CAN EXPRESS ANY ONE EQUATION AS A LINEAR COMINATION OF OTHER EQUATIONS ,THEN THEY ARE SAID TO BE LINEARLY DEPENDENT.
THAT IS SUPPOSE WE HAVE SEVERAL EQNS.LIKE
A1X+B1Y+C1Z+....ETC ...= 0..........1
A2X+B2Y+C2Z+....ETC... = 0.............2
A3X+B3Y+C3Z+....ETC... = 0...................3
-----------------------------------------------------------------------
-----------------------------------ETC-------------------------------
AND WE CAN WRITE .....
K1[A1X+B1Y+C1Z+....ETC..]+K2[2X+B2Y+C2Z+....ETC.]+K2[A3X+B3Y+C3Z+....ETC..] =0........................WITH NOT ALL K1,K2K3 ETC..ZERO ,
THEN WE SAY THEY ARE LINEARLY DEPENDENT.
IF WE CANNOT HAVE THE SUM AS ZERO UNLESS K1=K2=K3=...ETC=0 ALL , THEN WE SAY THEY ARE LINEARLY INDEPENDENT.

QUESTIONS - MAXIMA MINIMA

QUESTIONS - ANSWERS -LILLIAN - MAXIMA MINIMA

Blackwell, Lillian S. wrote:
>
>
>
>
> Could you please help me with the following problem. I tried to put it on
> the blog but I kept getting an error message.
>
>
>
> A dairy farmer plans to enclose a rectangular pasture adjacent to a river.
> The pasture must contain 180,000 square meters. No fencing is required
> along the river. What dimensions will use the smallest amount of fencing?

LET THE SIZE OF PASTURE BE
L IN LENGTH AND B IN WIDTH.
AREA = LB = 180000.........B = 180000/L
FENCING PERIMETER ON 3 SIDES = F = 2L+B = 2L + 180000/L
IF YOU KNOW DIFFERENTIATION,THEN
DF/DL = 2-180000/L^2=0
L^2 = 180000/2 = 90000
L=300 M
B=180000/300 = 600 M
------------------------------------------------------------
IF YOU DONT KNOW DIFFERENTIATION ,THEN..PUT L=X^2
F = 2[X^2+90000/X^2] = 2[(X^2)+(300/X)^2 - 2(X)(300/X) + 2(X)(300/X)]
F = 2[X - 300/X]^2 +1200
FOR F TO BE MINIMUM , X-300/X SHOULD BE ZERO.
X - 300/X =0
X^2 =300 = L
L=300
B = 180000/300 = 600

QUESTIONS - DIFFERENTIATION

QUESTIONS - ANSWERS - SHERILL - DIFFERENTIATION
sherill has left a new comment on your post "MATHS - 7.INDUCTION":
>
> Please Please Help! After T years, the value of a car purchased for $20,000
> is v=20,000(.75)^T When will the car be worth $500.00? Also, Use the
> derivative rule for y=b^x and find the derivative (rate of change) of the
> value of the car when: T=1 year and T=4 years

V = 20000*(0.75)^T=500
0.75^T = 500/20000=1/40
T*LOG 0.75 = -LOG(40)
T = -LOG(40)/LOG(0.75) = 12.823 YEARS.
DV/DT = 20000*(0.75^T)LOG(0.75)
WHEN T=1,WE GET
DV/DT = 20000*0.75*LOG(0.75)= -1874.1
WHEN T=4 , WE GET
DV/DT = 20000*(0.75^4)LOG(0.75)=-790.63

QUESTIONS - POLYNOMIALS - COMPLEX ROOTS

QUESTIONS-ANSWERS-POLYNOMIALS- COMPLEX ROOTS
On 11/13/06, Anonymous wrote:
> Anonymous has left a new comment on your post "MATHS - 7.INDUCTION":
>
> 1. For a polynomial f(x), suppose that f(12)=0. Then what is one factor of
> f(x)? 2. find a polynomial of degree 4 having zeros 2, i, -2i and

AS PER REMAINDER THEOREM IF F(12)=0 , THEN X-12 IS A FACTOR OF THE FUNCTION F(X).
IF THE POLYNOMIAL HAS REAL COEFFICIENTS THEN ..
IF I IS ZERO,THEN -I IS ALSO A ZERO
SIMILRLY IF -2I IS A ZERO,THEN 2I IS ALSO A ZERO
HENCE THE FACTORS ARE
(X-I),(X+I),(X-2I),(X+2I)....MULTIPLYING THESE FOUR FACTORS WE GET A 4 TH. DEGREE POLYNOMIAL
MULTIPLYING , WE GET
(X^2 - I^2)(X^2 - 4I^2) = (X^2+1)(X^2+4) = X^4+5X^2+4.=P(X)=0
IF X=2 IS ALSO A ZERO THEN WE SHALL NEED A 5 TH. DEGREE POLYNOMIAL...NOT 4 TH. DEGREE POLYNOMIAL.
(X-2)(X^4+5X^2+4) = X^5+5X^3+4X-2X^4-10X^2-8 = X^5-2X^4+5X^3-10X^2+4X-8 =F(X)=0

QUESTIONS - POLYNOMIALS COMPLEX ANALYSIS

QUESTIONS - ANSWERS -SUNNY- POLYNOMIALS - COMPLEX ANALYSIS

FIND IF (3-5i) IS A FACTOR OF
4X^4-28X^3+149X^2-70X-374.

USING REMAINDER THEOREM TO CHECK COMPLEX ROOTS IS OFTEN CUMBERSOME PARTICULARLY WITH LARGE COEFFICIENTS AS THIS AND 4 TH. DEGREE POLYNOMIAL
HENCE IT IS BETTER TO DO IT THE FOLLOWING WAY.
IF ….3-5i IS A FACTOR,THEN ITS CONJUGATE 3+5i SHOULD BE A FACTOR AS THE POLY NOMIAL HAS REAL COEFFICIENTS.
SO [X-(3-5i)][X-(3+5i)] SHOULD BE A FACTOR ..THAT IS
(X-3)^2 – (5i)^2 = X^2-6X+9+25 = X^2-6X+34 SHOULD BE A FACTOR. NOW DIVIDE THE POLYNOMIAL WITH THIS
X^2-6X+34] 4X^4 -28X^3 +149X^2-70X-374 [ 4X^2 – 4X –11
4X^4 –24X^3 +136X^2
-4X^3 + 13X^2 –70X
-4X^3 +24X^2 –136X
-11 X^2 +66X -374
-11X^2 +66X –374
0 REMAINDER
HENCE 3-5i IS A FACTOR.
THE REMAINING FACTORS ARE
X = [ 4 + OR – SQRT(4^2+4*4*11)]/(2*4)
= [4+ OR – 8 SQRT (3)]/8
X = (1/2) + OR – SQRT(3)
FACTORS ARE
[X – 0.5 – SQRT(3)] AND [X – 0.5 + SQRT(3)]

QUESTIONS - POLYNOMIALS

QUESTIONS - ANSWERS - POLYNOMIALS
On 11/12/06, Hope wrote:
> Hope has left a new comment on your post "Questions - Differential
> Equations":
>
> How do you do this? Divide x+3 into 2x^3+7x^2+x+3. Give the remainder and
> tell whether x+3 is a factor.
Remainder theorem helps to find remainder and hence whether a
particular binomial is a factor or not.But we cant get quotient from
it.
PUT X+3 =0..X=-3
DIVIDE USING SHORT DIVISION
-3|2 +7 +1 +3
0 -6 -3 +6
----------------------------
2 +1 -2 +9.............REMAINDER =9
QUOTIENT IS.... 2X^2 +X -2
X+3 IS NOT A FACTOR AS REMAINDER IS NOT ZERO.USING REMAINDER THEOREM
REMAINDER = F(-3) =
2*(-3)^3 +7(-3)^2+(-3)+3
=-54+63-3+3=9=REMAINDER


In the problem above, for what
> value of x does 2x^3+7x^2+x+3 equal 9? Find the solution w/o using direct
> substitution if you can, and explain your answer in terms of the remainder
> theorem.
AS WE SAW WITH REMAINDER THEOREM F(-3) = 9
SO FOR X=-3, WE GET F(-3) =9.
IF YOU DONT WANT DIRECT SUBSTITUTION, THEN THE DIVISION DONE ABOVE
GIVES F(-3) =9
BUT YOUR QUESTION SEEMS TO BE LITTLE ODD AND THE OBJECTIVE IS NOT CLEAR.

QUESTIONS - POLYNOMIALS - COMPLEX ROOTS

QUESTIONS - SUNNY - POLYNOMIALS - COMPLEX ROOTS
QUESTION BY SUNNY
WRITE AS PRODUCT OF 2 QUADRATIC FACTORS
ONE REDUCIBLE AND ANOTHER IRREDUCIBLE
THE POLYNOMIAL
P = X^4+6X^3-6X^2+6X-7
AS REASONED ABOVE POSSIBLE RATIONAL ROOTS ARE + OR - 1 AND 7 .
F(1)= 1+6-6+6-7 = 0….SO X=1 IS A ROOT..X-1=0…X-1 IS A FACTOR.
F(-7) = (-7)^4+6*(-7)^3-6*(-7)^2+6*(-7)-7=0
SO X= -7 IS A ROOT …X+7 = 0 ….X+7 IS A FACTOR
DIVIDING WITH THESE 2 FACTORS …

1) 1+6 -6 +6 -7
0+1+7+1+7
---------------------------
1 +7+1+7+0=REMAINDER
QUOTIENT = X^3 +7X^2 +X +7

-7) 1 +7+1+7
0 -7 +0 -7
------------------------
1 +0 +1 +0 = REMAINDER
QUOTIENT = X^2 +1
HENCE THE 2 QUADRATIC FACTORS ARE
(X^2+1) AND
(X-1)(X+7)= X^2+6X-7
ANOTHER METHOD IS TO LET
P = (X^2+AX+B)(X^2+CX+D)
AND SOLVING FOR A,B,C,D
COMPARING LIKE TERMS AND USING TRIAL & ERROR APPROACH
WE GET A=0,B=1,C=6 AND D=-7

QUESTIONS - POLYNOMIALS - COMPLEX ROOTS

QUESTIONS - SUNNY - POLYNOMIALS

QUESTION BY SUNNY ON POLYNOMIALS - COMPLEX ROOTS
FIND ALL ROOTS OF THE POLYNOMIAL-REAL AND COMPLEX
6X^5+13X^4-8X^3+16X^2-14X+3 = 0
IF FACTORS OF CONSTANT ARE C1,C2,C3 ETC…AND
FACTORS OF HIGHEST DEGREE TERM ARE H1,H2,H3 ETC..
THEN POSSIBLE RATIONAL ROOTS ARE + OR - C1/H1,C1/H2….C2/H1,C2/H2..ETC.
SO TRY RATIONAL ROOTS FIRST
6X^5+13X^4-8X^3+16X^2-14X+3
FACTORS OF CONSTANT TERM 3 ARE 1 AND 3.
FACTORS OF COEFFICIENT OF X^4 TERM - 6 - ARE 1,2,3 AND 6.
SO POSSIBLE RATIONAL ROOTS ARE + OR - 1,3,1/2,1/3,ETC
BY TRIAL WE FIND X= -3 IS A ROOT SINCE
6*(-3)^5+13*(-3)^4-8*(-3)^3+16*(-3)^2-14*(-3)+3 = 0
SO X =-3 THAT IS X+3=0 AND X+3 IS A FACTOR
DIVIDING WITH X+3
-3) 6 +13 - 8 + 16 -14 +3
0 -18 +15 -21 +15 -3
-----------------------------------
6 -5 +7 - 5 + 1 +0 REMAINDER
QUOTIENT = 6X^4 - 5X^3 + 7X^2 -5X +1

BY TRIAL WE FIND 1/2 AND 1/3 ARE ROOTS SINCE F(1/2) AND F(1/3) EQUAL ZERO
SO X=1/2 THAT IS 2X-1 = 0 AND HENCE 2X-1 IS A FACTOR.
SIMILARLY X=1/3 THAT IS 3X-1 = 0 AND HENCE 3X-1 IS A FACTOR.
DIVIDING SUCCESSIVELY WITH THESE 2 FACTORS

2X-1) 6X^4 - 5X^3 + 7X^2 -5X + 1 (3X^3 -X^2 +3X -1 =QUOTIENT
6X^4 -3X^3
-----------------------------------
-2X^3+7X^2
-2X^3+X^2
-----------------------------
6X^2 - 5X
6X^2 - 3X
-----------------
-2X +1
-2X +1
-----------------------
0

3X-1) 3X^3 - X^2 +3X -1 (X^2 + 1 = QUOTIENT
3X^3 - X^2
--------------------------
3X - 1
3X - 1
----------------------------
0

X^2+1 = (X-I) (X+I)…..THAT X = I AND -I ARE ROOTS.
SO ALL ROOTS ARE X = -3 , 1/2 , 1/3 ,I AND -I
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

QUESTIONS - INEQUALITIES & POLYNOMIALS

QUESTIONS - ANSWERS - INEQUALITIES
Anonymous has left a new comment on your post "Questions - Continuous Compounding":

Help please! Solve the following equation, then write solution in set notation. |x+10|<5



Posted by Anonymous to Maths,management,counselling at 8:57 AM

Reply Forward Invite Anonymous to Gmail

ANSWER:
|x+10|<5
case 1...
x+10>0....x>-10
|x+10|= x+10<5
x<5-10= -5
hence solution set is
(-10,-5)
case 2...
x+10<0....x<-10
|x+10|=-(x+10)=-x-10<5
-x<15
x>-15
hence solution set is
(-15,-10)


QUESTION - POLYNOMIALS

Anonymous has left a new comment on your post "Questions - Continuous Compounding":

OOps forgot one more. For the polynomial f(x)=-2x^3+3, which statement is true? a)as x goes to minus infinity, f(x) is increasing; and as x goes to infinity, f(x) is decreasing. b)as x goes to minus infinity, f(x) is decreasing; and as x goes to infinity, f(x) is decreasing c) neither Thank you!

ANSWER
f(x)=-2x^3+3
as x tends to plus infinity..x^3 tends to plus infinity and -x^3 tends to - infinity
hence f(x) tends to -infinity as x tend to infinity.That is f(x) is decreasing
similarly as x tend to - infinity , f(x) tends to plus infinity.That is increasing.

QUESTIONS - COMPLEX ANALYSIS

QUESTIONS - ANSWERS - COMPLEX ANALYIS
Question:
Please help~~
The function
f(z) = 1 / z^3
satisfies
integral f(z) dz =
for every closed curve C in the
c
complex plane that does not pass through the point 0. However, this function is not analytic at the point 0. Why does Morera's Theorem not apply?
Answer:
There are 2 stipulations to apply Morera's theorem.
1.f(z) should be continuous in a simply connected domain D.
2.Along any closed path in D integral f(z) should be zero.
Then the function will be anaytic in D.
In this case the second condition is satisfied while first condition is not satisfied as obviously the function is discontinuous at z=0 in D.
Hence,Morera's theorem is not applicable here.

QUESTIONS - POLYNOMIALS

QUESTIONS - ANONYMOUS - POLYNOMIALS
Anonymous has left a new comment on your post "PANCHATANTRA – 6.PROJECT MANAGEMENT":

Hello again, sir. I'm in need of your help. Determine the value of x where the following polynomial function touches the x-axis and tell whether it crosses the axis or is tangent to it. f(x)=-2x^3 Thank you.
f(x) = -2x^3 = 0 at x=0
f'(x) = -6x^2 = 0 at x =0
f"(x)= -12x = 0 at x= 0
f"'(x) = -12 = -12 at x=0
hence the curve crosses x axis at x=0 where y is also zero.
since its slope is zero at x=0, it means the slope of its tangent at x=0 is zero.
further it passes through x=0,y=0.hence
y-0 = 0*(x-0)=0 is the eqn. of the tangent at that point.
that is x axis is tangent to the curve at x=y=0.