2015/026) Given the 2 polynomials

$x^2 + ax + 1 = 0$
$x^2 + x + a = 0$

Find the value of a for which these have at least one root

solution

if both have a common root it shall satisfy the difference of the 2 polynomials

or ax – x + 1 – a = 0 or (x-1)(a-1) =0
so a = 1 or x = 1 giving a = -2

Q13/127) find integers in positive solutions to xyz + 20xy + yz + 5zx + 100x + 20y + 5z = 1913

Because we have xyz, xy, yz, xz, x, y, z terms in the left we should add some d and factor if possible

(x+a)(y+b)(z+c) – d

= xyz + ayz + bzx + cxy + bcx + abz + acy + abc – d

Comparing with given expression we get a = 1,  b = 5, c = 20

So we get (x+1)(y+5)(z+20) = xyz+ yz +20xy + 5xyz + 100 x + 5z + 20y + 100

Comparing with coefficient we get d = 100

So (x+1)(y+5)(z+20) = 1913 + 100 = 2013 = 3 * 11 * 61

So x = 2, y = 6, z= 41

2011/041) factoring by diminishing power $6bc - 9c² - 12cd - 8be + 12ce +16de$

one of the factoring that I saw in the net was factor

$6bc - 9c² - 12cd - 8be + 12ce +16de$

...the "clue" you need to recognize is that the coefficients of the first three terms have a common factor...and the coefficients of the last three terms have a common factor...and, the ratios are constant...so, group accordingly...

$(6bc - 9c^2 - 12cd) - (8be - 12ce - 16de) = 3c(2b - 3c - 4d) - 4e(2b - 3c - 4d) =$
$(2b - 3c - 4d)(3c - 4e)$

the above is correct and it was luck that grouping was there but what is luck is not there
then see that highest power of c is 2 and keep them in descending order

$ 6bc - 9c² - 12cd - 8be + 12ce +16de$
$= - 9c² + 6bc - 12cd + 12ce - 8be + 16de$
now you can factor as quadratic in c eliminating the luck factor
$= - 9c² - + 6c( b - 2d + 2e) -8e(b- 2d)$
$= - 9c² - + 6c(( b - 2d) + 2e)) -8e(b- 2d)$
letting b - 2d = a we get
$= - 9c^2 + 6c(a + 2e) - 8ea$
$= - 9c^2 + 6ca + 12ec - 8ea$
$= - 3c(3c - 2a) + 4e(3c-2a)$
$= (3c-2a)(4e - 3c)$
$= (3c-2b+4d)(4e-3c)$
which is same as 1st one

I do not mean to say that 2nd one is preferable to 1st but 2nd one can be used when 1st one does not work

2011/008) factor (a+b+c) ^3- a^3-b^3-c^3

we realize that

(a+b+c)^3-a^3 has a factor (b+c) and b^3 + c^3 has a factor b+c

so we proceed by combining

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

2010/053) If r and s are the roots of x^2 + x + 7 = 0, compute 2r^2 + rs + s^2 + r + 7

2r^2 + rs + s^2 + r + 7 =
r^2+rs + s^2 + (r^2+r+7)
= r^2+rs + s^2 as r^2+r+ 7 = 0 as r is a root
= (r+s)^2-rs ..1

now as x^2+x+7 has root r and s so rs = 7 and r+s = - 1

so from (1) we get given expression = 1-7 = - 6

2010/052) For which real values of p and q are the roots of x^3-px^2+11x-q=0 three consecutive integers

the three consecutive integers can be taken to be

s = a-1, r = a, t = a +1

coefficient of x = 11 = rs + rt + sr = a(a-1) + a(a+1) + (a-1)(a+1) = 3a^2 - 1 = 11

so a = 2 or -2
a=2 gives
roots are 1,2,3

so (x-1)(x-2)(x-3) = x^3- 6x^2 + 11x -6 or p = q = 6

a = -2 gives
roots as -1 ,-2,-3

(x+1)(x+2)(x+3) = x^3+6x + 11+ 6 so p=q = - 6

so 2 solutions are

1) p= q = 6
2) p =q = - 6

2010/045) If x + y = 2 and x^3+y^3 = 38,

then what does xy equal?

we know

(x+y)^3 = (x^3+y^3) + 3xy(x+y)

putting value from given conditions


2^3 = 38 + 3xy * 2 or 8 = 38 + 6xy => xy = - 5

2010/041) If (x-k)^2 is a factor of x^3 + 3px + q

prove that 4p^3 + q^2 =0
proof:
(x-k)^2 is a factor of x^3+3px + q so x-k is factor of this as well as derivative of this

so (x-k) is a factor so k^3 + 3pk + q = 0

and (x-k) is a factor of d/dx (x^3+3px + q) or 3x^2 + 3p or x^2+p =0

so k^2 + p = 0 ... 1 or p = -k^2

sum of zeros zero so 3rd factor = -2k

product = -2k^3 = q ...2

from 1 p = - k^2 or p^3 = -k^6 or 4p = -4k^6
and q^2 = 4k^6by adding we get 4p^3+q^2 = 0

2010/040) Let f(x) be a polynomial of degree>2

if f(x) is divided by (x-1) and(x-2)...?
it leaves a remainder 0,2 respectively.what is the remainder when f(x) is divided by (x-1)(x-2).

if f(x) is divided by (x-1)(x-2) remainder shall be a polynomial of degree 1

f(x) = p(x)(x-1)(x-2) + a x +b

f(1) = a + b = 0 so a = - b is the remainder when divided by (x-1)

f(2) = 2a + b = 2 so -2b +b = 2 or b - 2 and a = 2

so remainder = 2x - 2

2010/037) Sum of six consecutive whole squares can never be a whole square

let the 1st number be n^2

sum of 6 consecutive squares is

n^2+(n+1)^2+(n+2)^2+(n+3)^2+(n+4)^2 + (n+5)^2

= 6n^2+ 30n + 55
= 6n(n+5) + 55

1st part 6n(n+5)is divsible by 4 as either n or n+ 5 is even

so 6n(n+5) + 55 mod 4 = 55 mod 4 = 3

it cannot be a perfect square as perfect square mod 4 = 0 or 1

2010/024) If the roots of [b-c]x*x+[c-a]x+[a-b]=0are equal then prove that 2b=a+c

roots are equal so discriminant is zero

so (c-a)^2 - 4(b-c)(a-b) = 0

or c^2-2 ac + a^2 - 4(ab-ac- b^2 + bc) = 0

or a^2+4b^2+c^2+2ac - 4ab - 4bc = 0

or(a+c-2b)^2 = 0

or a+c-2b = 0 or a + c = 2b

proved

2010/005) If 0 < x < 1, then the sum of the infinite series (1/2)X^2 + (2/3)X^3 + (3/4)X^4 + ... is ...

let f(x) = (1/2)x^2 + (2/3)x^3 + (3/4)x^4 + ...

df/dx = x + 2 x^2 + 3x^3 + ... ....
= (x+x^2 + x^3...) + (x^2 + x^3 ...) + (x^3+x^4 + ....)
= x/(1-x) + x^2/(1-x) + ...
= x/(1-x)^2
= (1-x-1)/(1-x)^2
= 1/(1-x) - 1/(1-x)^2
integrating you get
f = log(1-x) + 1/(1-x) + constant of integration log(1-x) as 1- x > 0
this can be checked to be zero
so f(x) = 1/(1-x) + log(1-x)

2010/004) Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number.

Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number. Show that n?
is a prime.

Proof:
1 + x + x^2 + x^3 + ... + x^n-1 = (x^n-1)/(x-1)

now in case n is not a prime then say n= pq

now x^n-1 = (x^(pq)-1) = (y^q-1)/(x-1) where y = x^p
= (y-1)/(x-1)(y^(q-1) + y^(q-2) + ...1)
= (x^p-1)/(x-1) (y^(q-1) + y^(q-2) + ...1)
= (x^p-1 + +1)((y^(q-1) + y^(q-2) + ...1)
has got 2 factors and hence not a prime
so n must be prime

2010/003) The equation x^3 + 2x^ + 2x + 1 = 0 and x^200 + x^130 + 1 = 0 have ...?

exactly one common root;
b) no common root;
c) exactly three common roots;
d) exactly two common roots;
Kindly explain...

ans:
x^3 + 2x^2 + 2x + 1 = x(x+1)^2 + (x+1) = (x+1)(x^2 + x + 1) has zeros -1, w and w^2 where w is cube root of 1

-1 is not a zero of x^200 + x^130 + 1 since the expression equals 3 when x = -1.

So we are down to 2 possible roots in common.

let f(x) = x^200 + x^130 + 1

f(w) = w^200 + w^130 + 1= w^2+ w + 1 = 0 so w is a root
f(w^2) = w^400 + w^260 + 1 = w+w^2+1 = 0 so w^2 is a root

so there are exactly 2 common roots and hence d)

2010/002) The numbers 12n + 1 and 30n + 2 are relatively prime for ...?

any positive integer n;
b) infinitely many, but not all, integers n;
c) for infinitely many integers n;
d) none of the above;
Kindly explain...

Ans:
GCD(12n + 1 ,30n + 2)
= GCD(12n + 1 ,2(15n+1))
= GCD(12n + 1 ,15n+1) as 12n+1 is odd
= GCD(12n+1,(15n+1-(12n+1)) as GCD(a,b) = GCD(a,b-a)
= GCD(12n+1,3n)
= GCD(3n, 12n+1- 4*3n) as GCD(a,b) = GCD(b,a- n * b))
= GCD(3n,1)
= 1 for all n
hence ans is a

2009/029) prove a+b+c=0 ==> 2a⁴ +2b⁴ +2c⁴ = n²

a+b+c = 0

the a+b = - c
square both sides

(a^2+b^2+2ab) = c^2
so (a^2+b^2) = c^2 -2ab

now again square both sides

a^4+b^4 + 2a^2b^2 = (c^2-2ab)^2
or a^4 + b^4 + 2a^2b^2 = c^4 + 4a^2b^2 - 4abc^2
or a^4+b^4 = c^4 + 2a^2b^2 - 4abc^2
add c^4 on both sides to get

a^4+b^4+c^4 = 2c^4 + 2a^2b^2 - 4abc^2

or 2(a^4+b^4+c^4) = 4c^4+4a^2b^2 - 8abc^2
= 4(c^4-2abc^2+a^2b^2)
= 4(c^2-ab)^2

which is a perfect quare putting n = 2 | c^2-ab|
proved

2009/028) Rationalize the denominator of expression? [1 / {1 + x^1/5}]

Rationalize the denominator of expression?
[1 / {1 + x^1/5}]


ANS:
we know that (x^(1/5))^5 = x

say x^(1/5) = a

1+ x^ 1/5 = (1+ a)

as (1+a^5)/(1+a) = (1-a+a^2-a^3+a^4) and 1+a^5 in terms of x is rationalised

so
[1 / {1 + x^1/5}] = (1-x^(1/5) + x^(2/5) - x(^3/5) + x^(4/5))/(1+x)

2007/ 001) Slightly hard cubic eqution

Find constants a, b, c, and d such that:

4x³ - 3x + √3/2=a * (x-b) * (x-c) * (x-d)

Solution
let p(x) be given polynomial

we reallize that sqrt(3)/2 = cos pi/6

so 4x^3-3x + cos pi/ 6= 0

or - cos pi/6 =-3x + 4x^3 = cos(3t) if x= cos t

so cos 3t = - cos pi/6 = cos 5pi/6

so 3t = 5pi/6 or 2pi-5pi/6 = 7pi/6 or 4pi-7pi/6 = 17pi/6
so t = 5pi/18 or 7pi/18 or 17pi/18

so b = cos 5pi/18,
c = cos 7pi/18
d= cos 17pi/18

as these are zeroes of polynomials

further comparing coefficient of x^3 we get a = 4