2015/003) If $X$ and $Y$ are real root for the equations $x^3+3x^2+6x+20=0$ and $y^3+6y^2+15y-2=0$, find the sum of $X$ and $Y$

Now let
$P(x) = x^3+3x^2+6x+20$ and
$G(y) = y^3+6y^2+15y-2=0$

now let us check for p(x)

to eliminate the $x^2$ term I put x -1 to get
$P(x-1) = (x-1)^3 + 3(x-1)^2 + 6(x-1) + 20$
$= x^3-3x^2+ 3x - 1 + 3x^2 - 6x + 3 + 6x - 6 + 20$
$= x^3 + 3x + 16$

now let us check for g(y)
to eliminate the $y^2$ term I put y -2 to get
$G(y-2) =(y-2)^3+6(y-2)^2+15(y-2)-2$
$=y^3 - 6y^2 + 12y - 8 + 6y^ - 24y + 24 + 15y - 30 - 2$
$=y^3 + 3y - 16$

now if we put $R(x) = x^3 + 3x - 16$ then we get
P(x-1) = R(x) 
G(y-2) = - R(-y)

or P(x) = R(x+1) and G(y) = - R(y + 2)

so both are same and odd functions so
for the zeros of P and G the 2 values shall be -ve of each other hence

X + 1 = - (Y +2) or  X + Y  = - 3

I had solved incorrectly at http://mathhelpboards.com/challenge-questions-puzzles-28/find-x-y-13744.html#post65445.

2015/002) show that if $3x^2+x=4y^2+y$ then each of $x-y$ , $3x+3y+1$ and $4x+4y+1$ is a perfect square

we have

$3x^2+x = 4y^2 + y$
or $3x^2- 3y^2 + x - y = y^2$
or $3(x+y)(x-y) + (x-y) = y^2$
or $(3x+3y+1)(x-y) = y^2\cdots\, 1$
also
$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2\cdots\,2 $
multiply (1) and (2) to get

$(x-y)^2(3x+3y+1)(4x+4y+1)=x^2y^2$
or$(3x+3y+1)(4x+4y+1)=(\dfrac{xy}{x-y})^2$

so $(3x+3y+1)(4x+4y+1)$ is a perfect square

as $4(3x+3y+1) - 3(4x+4y+1) = 1$ so $(3x+3y+1)$ and $(4x+4y+1)$ both are perfect squares and the from (1) or (2) $(x-y)$ is a perfect square

2015/001) Find all solutions to $z^2 + 4z^* + 4 = 0$ where z is a complex number and $z^*$ is conjugate

Let z = a + ib

you get $(a+ib)^2 + 4(a-ib) + 4 = 0$

expand

$a^2 - b^2 + 2aib + 4a - 4bi + 4 = 0$
or
$(a^2 - b^2 + 4a + 4) + 2abi- 4bi = 0$

equate imaginary and real parts to zero to get

$(a^2 - b^2 + 4a + 4) = 0 \cdots 1$
and $2abi - 4bi = 0 => b = 0$ or a = 2

solve 1 using b = 0 to get a= - 2
sollve 1 using a= 2 to get b = $\pm4$

so (-2,0) is a solution so z = - 2
(2, 4i) and (2,-4i) are 2 other solutions so $z = 2 \pm 4i$ 

refer to https://in.answers.yahoo.com/question/index;_ylt=Aj51ZIERhFIw57.9V.6ln8.QHQx.;_ylv=3?qid=20120424152238AA6VkyD

2014/116) What is the value of m in the quadratic equation $x^2-mx+4=0$ to have real and equal roots.

equal roots

discriminant = 0 ($b^2-4ac, a = 1 b = -m . c = 4$)

$m^2 - 4*4 = 0$ or m= 4 or -4

m = 4 means $^2-4x+4 = 0$   oots are 2 and 2
m = -4 means $^2+4x+4 = 0$r  roots 2 and -2

roots are real as well


refer to https://in.answers.yahoo.com/question/index?qid=20111021024215AACQbhH

2014/115) how many factors of 240 are therer which are of the form 4n+ 2

240 can be factored as $2^4 * 3 * 5$

now for a factor to be of the form it should be 2 multiplied by by an odd number,
so it is 1 * (1+1) * (1+1) ( one way of choosing 2, 2 ways of choosing power of 3 ( $3^0, 3^1$) and same way 2 ways of choosing 5) or it is 4

2014/114) solve the equation $(x-3)^4 + (x-7)^4 = 24832$

Before I provide the solution I would like to mention that generally a quartic polynomial is not easy to solve but this type of equation

$(x-a)^4 + (x-b)^4= c$ can be converted to a quadratic equation by transformation of

$y= \dfrac{(x-a) + (x-b)}{2}$

as below

we shall put

$y= \dfrac{(x-3) + (x-7)}{2}$ = x- 5

so we get
$(y+2)^4 + (y-2)^4 = 24832$
or $2(y^4 + 6 y^2 (-2)^2 + 16) =  24832$
or $y^4 + 24 y^2 = 12400$
$y^4 + 24 y^2 – 12400 = 0$
 or $(y^2 – 100)(y^2 + 124)  = 0$
so $y^2 = 100$
so $y = \pm 10$

or x = -5 or 15

2014/113) find the number of solutions of $\sin\,x = \dfrac{x}{100}$

first let us look at positive x and same number of solutions shall be for -ve x

as $\sin,x$ and $x$ both are odd functions.

as $\sin\,x$ is less than 1 so x shall be less than 100 and so if we draw a sin  curve there shall be $\dfrac{100}{2\pi}$ or 15.91 ( around 16) units so there shall be 16 parts( 1/2 oscilaltions) in positive side and 16 in -ve side.

for each on the curve y = x shall intersect the sin curve 2 times so 32 in positive side including zero) so there is 1 value at 0, 31 positive values and 31 -ve values or 63 values