Omid Hatami wrote: Suppose $P,Q\in \mathbb R[x]$ that $deg\ P=deg\ Q$ and $PQ'-QP'$ has no real root. Prove that for each $\lambda \in \mathbb R$ number of real roots of $P$ and $\lambda P+(1-\lambda)Q$ are equal. deg(P)>1. First suppose P has at least one real root. Let r_1,...,r_n be the real roots of P (they are all distinct -of multiplicity 1- since if P(x)=0 then P'(x)<>0 because PQ'(x)-P'Q(x) <> 0). We can obviously assume deg(P)>=1, and since PQ'-P'Q has no real root, then (Q/P)'= (PQ'-P'Q)/P^2 keeps the same sign + or - in (r_i,r_i+1) for all i in [1,n-1]. Now we know that for any real root r_i of P: Q'(r_i)<>0 by hypothesis, and P changes sign at r_i ( because P'(r_i) <> 0 ); as a result we can pick a neighborhood V_i of r_i such that Q(V_i) does not contain {0} and P(r_i+) and P(r_i-) have opposite signs. From here you can find the possible variations of Q/P over R-{r_1,...,r_n}, and considering the fact that deg(P)=deg(Q), you have lim(P/Q, +oo) = lim(P/Q, -oo) = constant. There are only two possibilities for the variations of Q/P, and both of them will show you that the equation a*P+(1-a)*Q = 0 <=> Q/P = a/(a-1) ( the case a = 1 being trivial, and the case where P(x)=0 is ruled out since then Q(x)<>0 and the equation does not hold ) has exactly n solutions on R. If P has no real root then Q/P will be strictly monotous as above but lim(P/Q, -oo) = lim(P/Q, +oo), which is absurd (and hence P always has at least one real root). -- Julien Santini

Let's modify the conclusion to the statement that for all $t,\ Q+tP$ has the same number of real roots as $P$. Assume WLOG that $PQ'-QP'>0$. This says that the function $\frac QP$ is strictly increasing on every interval on which it's defined. Notice that $P(Q+tP)'-(Q+tP)P'>0$ as well in this case, so it suffices to prove the result for $t=0$ , i.e. it suffices to prove that $P$ and $Q$ have the same number of real roots. Let $a_1<\ldots<a_k$ be the distinct real roots of $P$. For $i\in\overline{1,k-1}$ the function $\frac QP$ goes to $-\infty$ as $x$ decreases to $a_i$ and to $\infty$ as $x$ increases to $a_{i+1}$, so $\frac QP$ and thus $Q$ has precisely one real root in each interval of the form $(a_i,a_{i+1})$. $\frac QP$ (and thus $Q$) also has precisely one root in exactly one of the intervals $(-\infty,a_1),(a_k,\infty)$, namely in the first one if the ratio of the dominant coefficients of $Q$ and $P$ is negative, and the second one if it's positive. This means that $\frac QP$ (and thus $Q$) has precisely $k$ real roots, as desired. Edit: Julien posted his message while I was typing mine.