**-------------**

**THEORY OF EQUATIONS:**

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(1) If an equation (i:e
f(x)=0 ) contains all positive co-efficients of any powers of x , it has no
positive roots then.

eg: x^4+3x^2+2x+6=0 has
no positive roots .

(2) For an equation , if
all the even powers of x have some sign coefficients and all the odd powers of
x have the opposite sign coefficients , then it has no negative roots .

(3)Summarising DESCARTES
RULE OF SIGNS:

For an equation f(x)=0 ,
the maximum number of positive roots it can have is the number of sign changes
in f(x) ; and the maximum number of negative roots it can have is the number of
sign changes in f(-x) .

Hence the remaining are
the minimum number of imaginary roots of the equation(Since we also know that
the index of the maximum power of x is the number of roots of an equation.)

(4) Complex roots occur
in pairs, hence if one of the roots of an equation is 2+3i , another has to be
2-3i and if there are three possible roots of the equation , we can conclude
that the last root is real . This real roots could be found out by finding the
sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
(More about finding sum and products of roots next time )

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07/10/2002 THEORY OF
EQUATIONS

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(1) For a cubic equation
ax^3+bx^2+cx+d=o

sum of the roots = - b/a

sum of the product of the
roots taken two at a time = c/a

product of the roots =
-d/a

(2) For a biquadratic
equation ax^4+bx^3+cx^2+dx+e = 0

sum of the roots = - b/a

sum of the product of the
roots taken three at a time = c/a

sum of the product of the
roots taken two at a time = -d/a

product of the roots =
e/a

(3) If an equation f(x)=
0 has only odd powers of x and all these have the same sign coefficients or if
f(x) = 0 has only odd powers of x and all these have the same sign

coefficients then the
equation has no real roots in each case(except for x=0 in the second case.

(4) Besides Complex roots
, even irrational roots occur in pairs. Hence if 2+root(3) is a root , then
even 2-root(3) is a root .

(All these are very
useful in finding number of positive , negative , real ,complex etc roots of an
equation )

Today's Section:

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08/10/2002

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(1) If for two numbers
x+y=k(=constant), then their PRODUCT is MAXIMUM if

x=y(=k/2). The maximum
product is then (k^2)/4 .

(2) If for two numbers
x*y=k(=constant), then their SUM is MINIMUM if

x=y(=root(k)). The
minimum sum is then 2*root(k) .

(3) |x| + |y| >= |x+y|
(|| stands for absolute value or modulus )

(Useful in solving some
inequations)

(4) Product of any two
numbers = Product of their HCF and LCM .

Hence product of two
numbers = LCM of the numbers if they are prime to each other .

1) For any regular
polygon , the sum of the exterior angles is equal to 360 degrees

hence measure of any
external angle is equal to 360/n. ( where n is the number of sides)

(2) If any parallelogram
can be inscribed in a circle , it must be a rectangle.

(3) If a trapezium can be
inscribed in a circle it must be an isosceles trapezium (i:e oblique sies
equal).

(4) For an isosceles
trapezium , sum of a pair of opposite sides is equal in length to the sum of
the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .

(5) Area of a regular
hexagon : root(3)*3/2*(side)*(side)

1) For any 2 numbers
a>b

a>AM>GM>HM>b
(where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa
respectively)

(2) (GM)^2 = AM * HM

(3) For three positive
numbers a, b ,c

(a+b+c) *
(1/a+1/b+1/c)>=9

(4) For any positive
integer n

2<= (1+1/n)^n <=3

(5) a^2+b^2+c^2 >=
ab+bc+ca

If a=b=c , then the
equality holds in the above.

(6) a^4+b^4+c^4+d^4
>=4abcd

(7) (n!)^2 > n^n (!
for factorial)

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This is for 21/10/2002

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

(1) If a+b+c+d=constant ,
then the product a^p * b^q * c^r * d^s will be maximum

if a/p = b/q = c/r = d/s
.

(2)Consider the two equations

a1x+b1y=c1

a2x+b2y=c2

Then ,

If a1/a2 = b1/b2 = c1/c2
, then we have infinite solutions for these equations.

If a1/a2 = b1/b2 <>
c1/c2 , then we have no solution for these equations.(<> means not equal
to )

If a1/a2 <> b1/b2 ,
then we have a unique solutions for these equations..

(3) For any quadrilateral
whose diagonals intersect at right angles , the area of the quadrilateral is

0.5*d1*d2, where d1,d2
are the lenghts of the diagonals.

(4) Problems on clocks
can be tackled as assuming two runners going round a circle , one 12 times as
fast as the other . That is ,

the minute hand describes
6 degrees /minute

the hour hand describes
1/2 degrees /minute .

Thus the minute hand
describes 5(1/2) degrees more than the hour hand per minute .

(5) The hour and the
minute hand meet each other after every 65(5/11) minutes after being together
at midnight.

(This can be derived from
the above) .

1)If n is even ,
n(n+1)(n+2) is divisible by 24

(2)If n is any integer ,
n^2 + 4 is not divisible by 4

(3)Given the coordinates
(a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting
point of the diagonals can be found out by solving for

[(a+e)/2,(b+f)/2] =[
(c+g)/2 , (d+h)/2]

(4)Area of a triangle

1/2*base*altitude =
1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where
s=a+b+c/2

=a*b*c/(4*R) where R is
the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle
.

(5)In any triangle

a=b*CosC + c*CosB

b=c*CosA + a*CosC

c=a*CosB + b*CosA

(6)If a1/b1 = a2/b2 =
a3/b3 = .............. , then each ratio is equal to

(k1*a1+
k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is
also equal to

(a1+a2+a3+............./b1+b2+b3+..........)

(7)In any triangle

a/SinA = b/SinB
=c/SinC=2R , where R is the circumradius

cosC = (a^2 + b^2 -
c^2)/2ab

sin2A = 2 sinA * cosA

cos2A = cos^2(A) - sin^2
(A)

1) x^n -a^n =
(x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding
multiples .For example (17-14=3 will be a multiple of 17^3 - 14^3)

(2) e^x = 1 + (x)/1! +
(x^2)/2! + (x^3)/3! + ........to infinity

(2a) 2 < e < 3

(3) log(1+x) = x -
(x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [ Note the alternating sign .
.Also note that the ogarithm is with respect to base e ]

(4) In a GP the product
of any two terms equidistant from a term is always constant .

(5) For a cyclic
quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where
s=(a+b+c+d)/2

(6) For a cyclic
quadrilateral , the measure of an external angle is equal to the measure of the
internal opposite angle.

(7) (m+n)! is divisible
by m! * n! .

_________________

"I have miles to go
before I sleep

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02/11/2002

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(1) If a quadrilateral
circumscribes a circle , the sum of a pair of opposite sides is equal to the
sum of the other pair .

(2)The sum of an infinite
GP = a/(1-r) , where a and r are resp. the first term and common ratio of the
GP .

(3)The equation whose
roots are the reciprocal of the roots of the equation

ax^2+bx+c is cx^2+bx+a

(4) The coordinates of
the centroid of a triangle with vertices (a,b) (c,d) (e,f)

is((a+c+e)/3 , (b+d+f)/3)
.

(5) The ratio of the
radii of the circumcircle and incircle of an equilateral triangle is 2:1 .

(6) Area of a
parallelogram = base * height

(7)APPOLLONIUS THEOREM:

In a triangle , if AD be
the median to the side BC , then

AB^2 + AC^2 = 2(AD^2 +
BD^2) or 2(AD^2 + DC^2) .

_________________

1) for similar cones ,
ratio of radii = ratio of their bases.

(2) The HCF and LCM of
two nos. are equal when they are equal .

(3) Volume of a pyramid =
1/3 * base area * height

(4) In an isosceles
triangle , the perpendicular from the vertex to the base or the angular
bisector from vertex to base bisects the base.

(5) In any triangle the
angular bisector of an angle bisects the base in the ratio of the

other two sides.

(6) the quadrilateral
formed by joining the angular bisectors of another quadrilateral is

always a rectangle.

(7) Roots of x^2+x+1=0
are 1,w,w^2 where 1+w+w^2=0 and w^3=1

( |a|+|b| = |a+b| if
a*b>=0

else |a|+|b| >= |a+b|

(9) 2<= (1+1/n)^n
<=3

(10) WINE and WATER
formula:

If Q be the volume of a
vessel

q qty of a mixture of
water and wine be removed each time from a mixture

n be the number of times
this operation be done

and A be the final qty of
wine in the mixture

then ,

A/Q = (1-q/Q)^n

(11) Area of a hexagon =
root(3) * 3 * (side)^2

(12) (1+x)^n ~ (1+nx) if
x<<<1

(13) Some pythagorean
triplets:

3,4,5 (3^2=4+5)

5,12,13 (5^2=12+13)

7,24,25 (7^2=24+25)

8,15,17 (8^2 / 2 = 15+17
)

9,40,41 (9^2=40+41)

11,60,61 (11^2=60+61)

12,35,37 (12^2 / 2 =
35+37)

16,63,65 (16^2 /2 =
63+65)

20,21,29(EXCEPTION)

(14) Appolonius theorem
could be applied to the 4 triangles formed in a parallelogram.

(15) Area of a trapezium
= 1/2 * (sum of parallel sids) * height = median * height

where median is the line
joining the midpoints of the oblique sides.

(16) when a three digit
number is reversed and the difference of these two numbers is taken , the
middle number is always 9 and the sum of the other two numbers is always 9 .

(17) ANy function of the
type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

(1 Let W be any point
inside a rectangle ABCD .

Then

WD^2 + WB^2 = WC^2 + WA^2

(19) Let a be the side of
an equilateral triangle . then if three circles be drawn inside

this triangle touching
each other then each's radius = a/(2*(root(3)+1))

(20) Let 'x' be certain
base in which the representation of a number is 'abcd' , then the decimal value
of this number is a*x^3 + b*x^2 + c*x + d

5) For a cyclic
quadrilateral , area = root( s* (s-a) * (s-b) * (s-c) * (s-d) ) , where
s=(a+b+c+d)/2

Here are some neat
shortcuts on Simple/Compound Interest.

Shortcut #1:

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

We all know the
traditional formula to compute interest...

CI = P*(1+R/100)^N - P

The calculation get very
tedious when N>2 (more than 2 years). The method suggested below is elegant
way to get CI/Amount after 'N' years.

You need to recall the
good ol' Pascal's Triange in following way:

Code:

Number of Years (N)

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

1 1

2 1 2 1

3 1 3
3 1

4 1 4
6 4 1

. 1 .... .... ... ... 1

Example: P = 1000, R=10
%, and N=3 years. What is CI & Amount?

Step 1: 10% of 1000 =
100, Again 10% of 100 = 10 and 10% of 10 = 1

We did this three times
b'cos N=3.

Step 2:

Now Amount after 3 years
= 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-

The coefficents - 1,3,3,1
are lifted from the pascal's triangle above.

Step 3:

CI after 3 years = 3*100
+ 3*10 + 3*1 = Rs.331/- (leaving out first term in step 2)

If N =2, we would have
had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs. 1210/-

CI = 2 * 100 + 1* 10 =
Rs. 210/-

This method is extendable
for any 'N' and it avoids calculations involving higher powers on 'N'
altogether!

A variant to this short
cut can be applied to find depreciating value of some property. (Example, A
property worth 100,000 depreciates by 10% every year, find its value after 'N'
years).

Shortcut #2:

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(i) When interest is
calculated as CI, the number of years for the Amount to double (two times the
principal) can be found with this following formula:

P * N ~ 72 (approximately
equal to).

Exampe, if R=6% p.a. then
it takes roughly 12 years for the Principal to double itself.

Note: This is just a
approximate formula (when R takes large values, the error % in formula
increases).

(ii) When interest is
calculated as SI, number of years for amt to double can be found as:

N * R = 100 . BTW this
formula is exact!

Adding to what 'Peebs'
said, this shortcut does work for any P/N/R.

Basically if you look
closely at this method, what I had posted is actually derived from the Binomial
expansion of the polynomial -- (1+r/100)^n but in a more "edible"
format digestable by us!

BTW herez one shortcut on
recurring decimals to fractions ...Its more easier to explain with an example..

Eg: 2.384384384 ....

Step 1: since the 3
digits '384' is recurring part, multiply 2.384 by 1000 = so we get 2384.

Next '2' is the non
recurring part in the recurring decimal so subtract 2 from 2384 = 2382.

If it had been
2.3848484.., we would have had 2384 - 23 = 2361. Had it been 2.384444.. NR
would be 2384 - 238 = 2146 and so on.

We now find denominator
part .......

Step 3: In step 1 we
multiplied 2.384384... by 1000 to get 2384, so put that first.

Step 4: next since all
digits of the decimal part of recurring decimal is recurring, subtract 1 from
step 3. Had the recurring decimal been 2.3848484, we need to subtract 10. If it
had been 2.3844444, we needed to have subtracted 100 ..and so on...

Hence here, DR = 1000 - 1
= 999

Hence fraction of the
Recurring decimal is 2382/999!!

Some more examples ....

1.56787878 ... = (15678 -
156) / (10000 - 100) = 15522/9900

23.67898989... = (236789
- 2367) / (10000 - 100) = 234422/9900

124.454545... = (12445 -
124) / (100 - 1) = 12321/99