If the question is not clear, here's an example. Let $p(x) := (x-1)(x-2) = x^2-3x+2$. Taking first and second derivatives, we get
\begin{align*}
p'(x) &= 2x -3
\end{align*}
and
\begin{align*}
p''(x) &= 2.
\end{align*}
Substituting these polynomials into the equation above we get
\begin{align*}
p''(x)p(x) – (p'(x))^2 &= 2x^2-6x+4 – (2x-3)^2.
\end{align*}
Which has roots $\frac{3}{2} \pm\frac{1}{2}i$ neither which are real. In the degree 2 case, it seems like $p''(x)p(x) – (p'(x))^2$ will always be another quadratic whose roots are complex, so it really comes down to showing that the discriminant is negative. Doing the computation for the polynomial
\begin{align*}
p(x) := (x-a)(x-b) = x^2-(a+b)x+ab
\end{align*}
we can call $c = a+b$ and computing derivatives we get
\begin{align*}
p'(x) &= 2x -c
\end{align*}
and
\begin{align*}
p''(x) &= 2.
\end{align*}. Substituting into our equation above, we get the simplification
\begin{align*}
-2x^2 + 2cx + (2ab-c).
\end{align*}
Using the quadratic formula, our roots are
\begin{align*}
x =\frac{-2c \pm \sqrt{2c^2-4(-2)(2ab-c)}}{-4}
\end{align*}
and substituting $a + b$ in for $c$, we get
\begin{align*}
x= \frac{a}{2} + \frac{b}{2} \pm \frac{\sqrt{2(a+b)^2-4(-2)(2ab-(a+b))}}{2}
\end{align*}
Which the real part of $x$ aligns perfectly with our example above. My issue is, how do I go about showing the discriminant is negative for arbitrary values $a$ and $b$? Furthermore, how do I show this generalizes to $n$ degrees?
Best Answer
Let $x_k$ be the real distinct roots, then $p(x) = a\prod_{k=1}^n (x-x_k)$ and:
$$ \frac{p'(x)}{p(x)} = \sum_{k=1}^n \frac{1}{x-x_k} \;\;\implies\;\; \frac{p''(x)p(x)-(p'(x))^2}{p^2(x)} = \left(\frac{p'(x)}{p(x)}\right)' = - \sum_{k=1}^n \frac{1}{(x-x_k)^2} \lt 0 $$
It follows that the LHS numerator $\,p''(x)p(x)-(p'(x))^2 \ne 0\,$ on $\,\mathbb R \setminus \{x_k\}\,$. Since all roots are distinct $\,p'(x_k) \ne 0\,$ so $\,p''(x)p(x)-(p'(x))^2 \ne 0\,$ on the whole $\,\mathbb R\,$.