Polynomial Root Existence
Find the smallest positive real number $ a $ so that the polynomial
\[x^6 + 3ax^5 + (3a^2 + 3) x^4 + (a^3 + 6a) x^3 + (3a^2 + 3) x^2 + 3ax + 1 = 0\]has at least one real root.
- 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
Note that $ x = 0 $ cannot be a real root. Dividing by $ x^3, $ we get
\[x^3 + 3ax^2 + (3a^2 + 3) x + a^3 + 6a + \frac{3a^2 + 3}{x} + \frac{3a}{x^2} + \frac{1}{x^3} = 0.\]Let $ y = x + \frac{1}{x} $. Then
\[y^2 = x^2 + 2 + \frac{1}{x^2},\]so $ x^2 + \frac{1}{x^2} = y^2 - 2, $ and
\[y^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3},\]so $ x^3 + \frac{1}{x^3} = y^3 - 3y $. Thus,
\[y^3 - 3y + 3a (y^2 - 2) + (3a^2 + 3) y + a^3 + 6a = 0.\]Simplifying, we get
\[y^3 + 3ay^2 + 3a^2 y + a^3 = 0,\]so $ (y + a)^3 = 0 $. Then $ y + a = 0, $ so
\[x + \frac{1}{x} + a = 0.\]Hence, $ x^2 + ax + 1 = 0 $. For the quadratic to have real roots, the discriminant must be nonnegative, so $ a^2 \ge 4 $. The smallest positive real number $ a $ that satisfies this inequality is $ a = \boxed{2} $.
Freeflows