<s>[INST] If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ be a polynomial where a and b are real numbers and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a.
Since the function has equal positive roots, I expect $f'(\alpha) = 0$ and $f''(\alpha) > 0$ for the positive root $\alpha$ But I'm not sure how I should proceed in this question. What are the different methods of doing this problem?
 [/INST] Note that if $r\neq 0$ is a root of $f(x)=0$, then so is $\frac{1}{r}$. This is easily noted by the symmetry of the coefficients and can algebraically shown by the fact that
$$r^4f(\frac{1}{r})=f(r)$$
Hence, if the positive root is $r$ and the negative root of smaller magnitude is $s$, then our four roots are
$$r,\frac{1}{r},s,\frac{1}{s}$$
However, since $r$ and $\frac{1}{r}$ are given as equal positive numbers, we have $$r=1$$
Hence, $f(x)$ is of the form
$$f(x)=(x-1)^2(x-s)(x-\frac{1}{s})$$
$$f(x)=(x^2-2x+1)(x^2-(s+\frac{1}{s})x+1)$$
$$f(x)=x^4-(2+s+\frac{1}{s})x^3+(2+2s+\frac{2}{s})x^2-(2+s+\frac{1}{s})x+1$$
Hence, for all $s\neq 0$ we have that $f(x)$ fits the form $x^4+ax^3+bx^2+ax+1$.
Note that the value of $a$ is
$$a=-2-s-\frac{1}{s}$$
$$a=-2+(-s)+\frac{1}{(-s)}$$
Since $s$ is given as a negative number, we have that $(-s)$ must be a positive number. Hence, by AM-GM we have
$$\frac{(-s)+\frac{1}{(-s)}}{2}\geq \sqrt{(-s)\left(\frac{1}{(-s)}\right)}$$
$$(-s)+\frac{1}{(-s)}\geq 2$$
with equality when $(-s)=\frac{1}{(-s)}$. This occurs when $-s=1$. However, since this results in both negative roots being the same, equality is never attained.
Moreover, we have
$$a=-2+\left((-s)+\frac{1}{(-s)}\right)\geq -2+2$$
$$a\geq 0$$
As we showed before, this equality is unattainable. The next integral value of $a$ is $\boxed{1}$. This is attainable when
$$(-s)+\frac{1}{(-s)}=3$$
$$(-s)^2-3(-s)+1=0$$
$$(-s)=\frac{3\pm\sqrt{5}}{2}$$
Note that the $2$ solutions are positive, and will be the solutions to $(-s)$ and $\frac{1}{(-s)}$. Since we defined $s$ as the root with smaller magnitude, we have $s=-\frac{3-\sqrt{5}}{2}$.
Hence, the equality of $\boxed{a=1}$ is attained when
$$f(x)=(x^2-2x+1)(x^2+3x+1)$$
$$f(x)=x^4+x^3-4x^2+x^3+1$$
</s>