Cubic Polynomial Property
A cubic polynomial $ f(x) = x^3 + ax^2 + bx + c $ with at least two distinct roots has the following properties.
(i) The sum of all the roots is equal to twice the product of all the roots.
(ii) The sum of the squares of all the roots is equal to 3 times the product of all the roots.
(iii) $ f(1) = 1 $.
Find $ c $.
- 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 $ r, $ $ s, $ $ t $ be the root of the cubic. Then by Vieta's formulas,
\begin{align*}
r + s + t &= -a, \\
rs + rt + st &= b, \\
rst &= -c.\end{align*}From condition (i), $ -a = -2c, $ so $ a = 2c $.
Squaring the equation $ r + s + t = -a, $ we get
\[r^2 + s^2 + t^2 + 2(rs + rt + st) = a^2.\]Then
\[r^2 + s^2 + t^2 = a^2 - 2(rs + rt + st) = a^2 - 2b.\]Then from condition (ii), $ a^2 - 2b = -3c, $ so
\[b = \frac{a^2 + 3c}{2} = \frac{4c^2 + 3c}{2}.\]Finally, from condition (iii), $ f(1) = 1 + a + b + c = 1, $ so $ a + b + c = 0 $. Substituting, we get
\[2c + \frac{4c^2 + 3c}{2} + c = 0.\]This simplifies to $ 4c^2 + 9c = 0 $. Then $ c(4c + 9) = 0, $ so $ c = 0 $ or $ c = -\frac{9}{4} $.
If $ c = 0, $ then $ a = b = 0, $ which violates the condition that $ f(x) $ have at least two distinct roots. Therefore, $ c = \boxed{-\frac{9}{4}} $.
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