What is the sum of X and Y for real roots of two equations?

[anemone]

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 $Y$.

 

[kaliprasad]

Spoiler

I had solved a similar problem ( but one equation) at http://mathhelpboards.com/challenge-questions-puzzles-28/find-m-n-11000.html#post51080.
I shall use the same approach to see if they are odd functions and same with a little transformationnow 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
$G(y-2) =(y-2)^3+6(y-2)^2+15(y-2)-2$
$=y^3 - 6y^2 + 12y - 8 + 6y^2 - 24y + 24 + 15y - 30 - 2$
$=y^3 + 3y - 16$now if we put $R(x) = x^3 + 3x - 16$ then we get

P(x-1) = R(x)
G(y-2) = - R(-y)

so both are same function 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 2 - Y or X + Y = 3

 

[anemone]

Good job, kaliprasad!

I wasn't sure at first if I should post this challenge, knowing it is quite similar to that one you cited (which is also a challenge problem that I posted), but I went ahead and post it, with the hope that no one would remember it, not even you or Albert...hehehe...

Anyway, thanks for participating again of this challenge, my friend!:)

 

[Albert1]

Spoiler

I had solved a similar problem ( but one equation) at http://mathhelpboards.com/challenge-questions-puzzles-28/find-m-n-11000.html#post51080.
I shall use the same approach to see if they are odd functions and same with a little transformationnow 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
$P(x-1) = (x-1)^3 + 3(x-1)^2 + 6(x-1) + 20$
$= x^3-3x^2+ 3x - 1 + 3x^2 - 6x + 3 + 6x - 6 + 20$
$= x^3 + 3x + 16----(1)$now let us check for G(y)
to eliminate the $y^2$ term I put y -2 to get
$G(y-2) =(y-2)^3+6(y-2)^2+15(y-2)-2$
$=y^3 - 6y^2 + 12y - 8 + 6y^2 - 24y + 24 + 15y - 30 - 2$
$=y^3 + 3y - 16$now if we put $R(x) = x^3 + 3x - 16--(2)$ then we get

P(x-1) = R(x) ---(3)
G(y-2) = - R(-y)---(4)

so both are same function 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 2 - Y or X + Y = 3

mistakes found ,please check (1)(2)(3)(4)
and the answer

 

[kaliprasad]

mistakes found ,please check (1)(2)(3)(4)
and the answer

right and thanks

Spoiler

R(x) should be $x^3+3x + 16$

 

[Albert1]

x+y should be -3

 

[kaliprasad]

x+y should be -3

I give up. I would like someone to correct this part.

 

[Albert1]

$P(x)=R(x+1)=0$
$G(y)=-R(-y-2)=0$
$\therefore x+1=-y-2$
$x+y=-3$

 

[kaliprasad]

$P(x)=R(x+1)=0$
$G(y)=-R(-y-2)=0$
$\therefore x+1=-y-2$
$x+y=-3$

Thanks for correcting me. It was a silly mistake by me

 

[anemone]

I want to apologize to MHB, kaliprasad and Albert for not checking the answer before jumping to a conclusion, and I also want to thank Albert for his "eagle eye" in spotting the error so that the solution is deemed to be a good one now. (Whew):)

Way to go the extra mile, Albert!(Nod)