Let a, b, and c be the roots of the polynomial p(x)=x^3+2x^2+3x+4.

Let g(x) be the monic polynomial?

whose roots are a+b, a+c, and b+c (each with multiplicity 1). Determine g(x).

sum of the roots = - coefficient of x^2

hence

a + b + c = -2

So a+b = -2-c

a +c = - 2-b

b+c = - 2- a

so we need to find equation whose roots are 2-a, 2-b 2-c

p(x)=x^3+2x^2+3x+4 has roots a,b,c

so p(-x) = -x^3 + 2x^2 – 3x + 4

or f(x) = x^3 – 2x + 3x – 4 (dividing p(-x) by – 1) has roots –a , -b ,- c

so f(x+2) = (x+2)^3 – 2(x+2)^2 + 3(x+2) – 4 has roots -2 – a , -2 –b , -2 - c

so g(x) = f(x+ 2) = x(x+2)^2 + 3 x+ 6- 4

= x(x^2 + 4x+ 4)+ 3x + 6 -2

= x^3 + 4x^2 + 7x + 4 is the required result.

If x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b)) … then prove that bx2-ax+b=0

x =(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))

use Componendo and dividendo

(x+1)/(x-1) = √(a+2b)/√(a-2b)

square both sides

(x^2 + 2 x + 1)/(x^2-2x + 1) = (a+2b)/(a-2b)

again use Componendo and dividendo

(x^2 + 1)/(2x) = (a)/(2b)

or bx^2 +b = ax
or bx^2 - ax + b = 0

Note: other method can be used but it is simpler when we have form (a+b)/(a-b)

Source(s):

http://en.wikipedia.org/wiki/Componendo_…

Let x = ln(sec(y) + tan(y). Show that sec(y) = cosh (x).?

Remember that cosh(x) = (1/2) * (e^(x) + e^(-x)) ..1

and as sec^2 y- tan ^2 y = 1 so 1/(sec y + tan y) = sec y - tan y ..2

x = ln(sec(y) + tan(y))

so e^x = sec(y) + tan(y) ...3
so e^-x = 1/( sec y + tan y) = sec y - tan y .. 4 ( from 2)

add (3) and (4) to get e^x + e^-x = 2 sec (y)

or sec y = (e^ x + e^-x)/2 = cosh x (from 1)

prove (tan12)(tan48)(tan54)(tan72) = 1

we have
tanθ tan(60° - θ) tan (60° + θ) = tan3θ. (proof is below)

put θ =12 to get
tan 12 tan(60 - 12) tan (60 + 12) = tan 3 * 12
=> tan 12 tan 48 tan 72 = tan 36
=> tan 12 tan 48 tan 72 = cot (90-36) = cot 54 = 1/ tan 54
=> (tan12)(tan48)(tan54)(tan72) = 1
proved
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to prove

tanθ tan(60° - θ) tan (60° + θ) = tan3θ

tan(60° - θ) tan (60° + θ)
= ((tan 60 - tan θ)/(1+ tan 60 tan θ))((tan 60 - tan θ)/(1+ tan 60 tan θ))
= (tan^2 60 - tan^2 θ)/(1- tan^2 60 tan^2 θ))
= (3 - tan θ)/(1 - 3 tan^2 θ))

so tanθ tan(60° - θ) tan (60° + θ) = (3 tan θ - tan^3 θ)/(1 - 3 tan^2 θ)) = tan 3θ

proved

If p and q are distinct primes and x^2 - px + q = 0 has distinct positive integral roots, then p + q has value equal to

it is got positive integral roots so roots are 1 and q as q is prime

equation is (x-1)(x-q)
= x^2 - (q+1) + q = 0

so p = q + 1

so q and q +1 are primes so q = 2 and q +1 =3 so p = 3, q = 2 and p+q = 5

On squared paper, draw a rectangle and one of its diagonals. How many grid squares are crossed by the diagonal?

Let a = (0,0) and x coordinate is to right and y coordinate downwards

Let the rectangle be ABCD as in the diagram and of the size m * n. The diagonal is AC. There are m+ 1 vertical lines and n+1 horizontal in the above diagram m = 4 and n = 3

Now let us see when it shall pass though a point which is a corner that is intersection of horizontal line and vertical line say at (p , q) coordinate

Slope of the line = m/n = p/ q where p < m , q < n .

This may pass through multiple points and let p/q is with the lowest p and q

m/p = n/q is the GCD (m,n)

So if gcd(m,n) is 1 then we do not have p/q form and the line does not pass through any point (p,q) that is a corner( we define a corner as intersection of horizontal line and vertical line)

So we take 2 cases

1) GCD(m,n) is 1 that is m,n are coprime

The diagonal from A to C shall pass through m vertical sections which shall be m squares and n-1 horizontal lines shall be cut by the diagonal ( at a point other than a corner point) so each shall give 2 squares that is addition of 1 square that making m+n-1 squares. So the diagonal pass through m+n-1 squares.

2) GCD(m,n) is not one that is m ,n are not coprimes say GCD (m,n) = p

Now m/p and n/p are coprimes and we get p parts of (m/p, n/p) rectangle through which the diagonal passes.

So number of points = p(m/p + n/p – 1) = m+ n – p

So we combining (1) and (2) get m+ n – gcd(m,n)

For example in the above figure m = 4, n= 3.

It goes through 1+ 2 + 2 + 1 ( note that in the 2^nd region and 3^rd region it is 2 as diagonal cuts the horizontal line) = 6

m+n – 1 = 4 + 3 -1 = 6

prove that tan20+4sin20=root 3

tan20+4sin20
= sin 20/ cos 20 + 4 sin 20
= ( sin 20 + 4 sin 20 cos 20) / cos 20
= (sin 20 + 2 sin 40)/ cos 20
= (sin 20 + 2 sin (60-20))/ cos 20
= ( sin 20 + 2 sin 60 cos 20 - 2 cos 60 sin 20)/ cos 20
= (sin 20+ 2 sin 60 cos 20 - sin 20)/ cos 20
= 2 sin 60cos 20/ cos 20
= 2 sin 60
= 2(√3/2) = √3

alternatively


we can proceed from (sin 20 + 2 sin 40)/ cos 20

as ( sin 20 + sin 40 + sin 40) / cos 20
= (2 sin 30 cos 10 + sin 40)/ cos 20
= (cos 10 + sin 40)/ cos 20
= ( sin 80 + sin 40)/ cos 20
= 2 sin 60 cos 20/ cos 20
= 2 sin 60
= √3