Real Triplets Count
Find the number of ordered triples $ (x,y,z) $ of real numbers such that
\begin{align*}
x + 2y + 4z &= 12, \\
xy + 2xz + 4yz &= 22, \\
xyz &= 6.\end{align*}
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ a = x, $ $ b = 2y, $ and $ c = 4z $. Then $ x = a, $ $ y = \frac{1}{2} b, $ and $ z = \frac{1}{4} c, $ so the given system becomes
\begin{align*}
a + b + c &= 12, \\
ab + ac + bc &= 44, \\
abc &= 48.\end{align*}Then by Vieta's formulas, $ a, $ $ b, $ and $ c $ are the roots of
\[t^3 - 12t^2 + 44t - 48 = 0.\]This factors as $ (t - 2)(t - 4)(t - 6) = 0, $ so $ a, $ $ b, $ $ c $ are 2, 4, 6, in some order.
There are $ 3! = 6 $ ways to assign 2, 4, 6 to $ a, $ $ b, $ and $ c $. These produce $ \boxed{6} $ different solutions $ (x,y,z), $ via the substitution $ x = a, $ $ y = \frac{1}{2} b, $ $ z = \frac{1}{4} c $.
Freeflows