Consider all quadruples of real numbers (x1,x2,x3,x4), such that x1≤x2≤x3≤x4 and each of the four numbers plus the product of the other three equals 130. If S is the sum of all possible values of x4, what is S,rounded to the nearest integer?
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Consider all quadruples of real numbers (x1,x2,x3,x4), such that x1≤x2≤x3≤x4 and each of the four numbers plus the product of the other three equals 130. If S is the sum of all possible values of x4, what is S,rounded to the nearest integer?
1,2,3,4 are the subscripts of x
Hello, geniusgarvil!
Quote:
Consider all quadruples of real numbers $\displaystyle a,b,c,d$ , such that $\displaystyle a \le b \le c \le d$
and each of the four numbers plus the product of the other three equals 130.
If $\displaystyle S$ is the sum of all possible values of $\displaystyle d$, what is $\displaystyle S$? .(Round to nearest integer.)
We have: .$\displaystyle \begin{Bmatrix}a + bcd \:=\:130 \\ b + acd \:=\:130 \\ c + abd \:=\:130 \\ d + abc \:=\:120 \end{Bmatrix}$
Since the four variables are completely interchangeable,
. . we conclude that: .$\displaystyle a \,=\, b \,=\, c \,=\, d.$
The first equation becomes: .$\displaystyle a + a^3 \:=\:130 \quad\Rightarrow\quad a^3 + a - 130 \:=\:0$
. . The only real root is: .$\displaystyle a \,=\,5.$
Therefore: .$\displaystyle a \,=\,b\,=\,c\,=\,d\,=\,5 \quad\Rightarrow\quad S \,=\,5$
I,don't know what's wrong in your solution, but the answer's incorrect.
