<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] You have
$$\begin{aligned}
f(x)&=x^4+ax^3+bx^2+ax+1\\
&= x^2\left(x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}\right)\\
&=x^2\left(\left(x + \frac{1}{x}\right)^2+ a\left(x + \frac{1}{x}\right)+b-2\right)
\end{aligned}$$
As $x=0$ is not a root of $f(x)$, $x$ is a root of $f(x)$ if and only if $X =x + \frac{1}{x}$ is a root of $g(X)= X^2+a X + b-2$. As all roots of $f$ are supposed to be real, so are the ones of $g$. Hence $\Delta = a^2 + 8-4b \gt0$. The roots of $g$ are
$$X_1 = \frac{-a - \sqrt{\Delta}}{2} \lt 0 \text{ and } X_2 = \frac{-a + \sqrt{\Delta}}{2} \gt 0$$
If $x_1 \lt x_2 \lt 0 \lt x_3=x_4$ are the roots of $f$, we have $X_2 = 2$ as $2$ is the only value for which the equation $x + \frac{1}{x}=X_2$ has a unique positive root. Which implies
$$a^2 + 8-4b = (4+a)^2 = 16+8a+a^2 \iff 2a+b+2=0, $$ hence $$\Delta=a^2+8-4b= a^2+8+8a+8=a^2+8a+16.$$ And as the negative roots of $f$ are distinct$$X_1 = \frac{-a - \sqrt{\Delta}}{2}\lt -2.$$ Which leads to
$$a^2+8a+16 \gt (4-a)^2$$ i.e.
$$16a \gt 0$$ i.e. $a \gt 0$. The minimum integral value for $a$ is therefore $1$. In that case, $$\Delta = 25, X_2=2, x_3=x_4=1$$ and $x_1,x_2$ are the values for which $$x + \frac{1}{x} = -3.$$
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