Prove that the expression x^2 - (a+b+c)x+a^2+b^2+c^2+2bc -ca-ab can never be negative if x,a,b,c, are real.

There are two proofs to it

1) let us multiply by 4 to make it denominator free

4x^2 - 4(a+b+c)x+4(a^2+b^2+c^2+2bc -ca-ab)

= (2x - (a+b+c))^2 + 4(a^2+b^2+c^2+2bc -ca-ab) - (a+b+c)^2

= (2x - (a+b+c))^2 + 4(a^2+b^2+c^2+2bc -ca-ab) - (a^2+b^2 + c^2 + 2ab + 2bc + 2ca)

= (2x - (a+b+c))^2 + (3a^2+3b^2+3c^2+6bc -6ca-6ab)

= (2x - (a+b+c))^2 + 3(a^2+b^2+c^2+2bc -2ca-2ab)

= (2x - (a+b+c))^2 + 3(a^2+(b+c)^2 -2a(b+c))
= (2x - (a+b+c))^2 + 3(a-(b+c))^2

as it it sum of squares it cannot be -ve

Method 2

Using the quadratic formula gives:

x = (-(-(a+b+c) ± √((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)))/2

So if ((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)) < 0 it will only have complex roots and so wont cross the x axis. Since the coefficient of x² is positive this means it would only be positive. Therefore one need to prove ((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)) < 0

o r

-3a² -3b² - 3c² + 6ab + 6ac - 6bc < 0

factoring gives
-3(a-b-c)² < 0

which is true unless a-b-c = 0 it will have complex roots and so only be positive.

If a-b-c = 0 that means it has equal roots and so will only touch the x axis without crossing it and so will only be positive or 0.

Prove that n^5 - n is divisible by 30

proof:
we have n^5 - n = n(n^4-1) = n(n^2-1)(n^2+1) = n(n+1)(n-1) (n^2+1)

as n(n+1)(n-1) is product of 3 consecutive numbers it is divisible by 6

now under mod 5
n^2 + 1 = n^2- 4 = (n+2)(n-2)
so n^5 - n = n(n+1)(n-1)(n+2)(n-2)

as it is product of 5 consecutive numbers it is divisible by 5

as it is divisible by 5 and 6 hence 30

proved

solve (x+2)(x+3)(x+8)(x+12)=4x^2?

we see 2 * 12 = 3 * 8 = 24

so group as follow

(x+2)(x+12)(x+3)(x+8)=4x^2

=> (x^2 + 14 x + 24) (x^2 + 11x + 24) = 4x^2
let x^2 + 11x +24 be y
(y+ 3x) y = 4x^2
y^2 + 3xy - 4x^2 = 0
(y- x)(y+ 4x) = 0

y =x => x^2 + 11x + 24 = x => x^2 + 10x + 24 = 0 => (x+4)(x+6) = 0 => x = -4 or - 6

y+ 4x = 0 => x^2 + 15x + 24 = 0

this can be solved for complex solutions

x = -15/2 - SQRT(129)/2 or x = -15/2 + SQRT(129)/2

Show 1 + cisx + cis2x = (1-cis3x)/(1-cisx)?

we know
cis x = cos x + i sin x

(cis x)^n = cos ^n x + i sin ^n x = cis nx as per http://en.wikipedia.org/wiki/De_Moivre%2…


so LHS = 1 + cisx + cis2x = 1 + cis x + cis ^2 2x

let cis x = t (not required but makes understanding easier)

LHS = 1 + t + t^2 = (1-t^3)/(1- t) = (1- cis ^3 x)/(1- cis x) =
= ( 1- cis 3x )/(1- cis x) = RHS

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