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anemone
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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 $Y$.
mistakes found ,please check (1)(2)(3)(4)kaliprasad said: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
Albert said:mistakes found ,please check (1)(2)(3)(4)
and the answer
Albert said:x+y should be -3
Albert said:$P(x)=R(x+1)=0$
$G(y)=-R(-y-2)=0$
$\therefore x+1=-y-2$
$x+y=-3$
"Find the Sum of X+Y" is a mathematical expression that requires you to add together the values of X and Y to determine their total sum.
The purpose of finding the sum of X+Y is to solve a mathematical problem or equation that involves adding two variables together.
To find the sum of X+Y, you will need to add the value of X to the value of Y. For example, if X=5 and Y=3, the sum of X+Y would be 5+3=8.
The correct order of operations when finding the sum of X+Y is to first add the values of X and Y, and then solve any remaining operations in the equation from left to right.
There are no special cases when finding the sum of X+Y. This mathematical operation follows the basic principles of addition and can be applied to any values of X and Y.