please help me with this..
Solve for the set of equations :
$\displaystyle x+y+z=12 \\ xy+yz+zx=44 \\ xyz=48 $
And...Merry Christmas & Happy new Year
Merry Christmas! ^^
$\displaystyle xyz=48\Rightarrow x\neq 0, y\neq 0, z\neq 0$, knowing this we can multiply/ divide by x, y or z.
Starting from the first relation, we have:
$\displaystyle x+y+z=12\Rightarrow x(x+y+z)=12x\Rightarrow x^2+xy+xz=12x\Rightarrow$
$\displaystyle \Rightarrow x^2+xy+xz-(xy+yz+zx)=12x-44\Rightarrow x^2-yz=12x-44$
We know that $\displaystyle xyz=48$, so that, yz which is kind of a problem for us, can be written as $\displaystyle \frac{48}{x}$:
$\displaystyle x^2-\frac{48}{x}=12x-44 \Rightarrow x^3-48=12x^2-44x\Rightarrow x^3-12x^2+44x-48=0$
For x=2: $\displaystyle 2^3-12\cdot 2^2+44\cdot 2-48=8-48+88-48=0$. That means that we write our equation this way:
$\displaystyle x^3-12x^2+44x-48=x^3-2x^2-10x^2+20x+24x-48=$
$\displaystyle =x^2(x-2)-10x(x-2)+24(x-2)=(x-2)(x^2-10x+24)=0$
So we know that: $\displaystyle (x-2)(x-4)(x-6)=0$. x can then be 2, 4 or 6.
We can (and actually should) do the same thing for y and z, so y and z can also be 2, 4 or 6.
$\displaystyle (x, y, z) \in \left \{ (2, 4, 6); (2, 6, 4); (4, 2, 6); (4, 6, 2), (6, 2, 4); (6, 4, 2) \right \}$
Hello, earthboy!
$\displaystyle \text{Solve: }\;\begin{Bmatrix}x+y+z\;=\;12 \\ xy+yz+zx\;=\;44 \\ xyz\;=\;48 \end{Bmatrix}$
$\displaystyle \text{We note that }x,y,z\text{ are the roots of the equation:}$
. . . $\displaystyle u^3 - 12u^2 + 44u - 48 \:=\:0$
$\displaystyle \text{Factor: }\:(u-2)(u-4)(u-6) \:=\:0$
$\displaystyle \text{Therefore: }\:\{x,y,z\} \:=\:\{2,4,6\}\text{ in some order.}$