Any Real?

As said in the question that a and b are roots of $P(x)$ and $Q(x)$ respectively then it can be said that $P(a)=0$ and $Q(b)=0$ .

Let, consider a new polynomial such that $R(x)=P(x)-Q(x)$ .

Put $a$ and $b$ in this equation and we get
$R(a)=P(a)-Q(a)=-Q(a)$ and $R(b)=P(b)-Q(b)=P(b)$ .

Now, $R(a)R(b)= -P(b)Q(a) <0$ because $P(b)Q(a)>0$ in the question

So, there exists a real $c$ in between $(a,b)$ such that $R(c)=0$ because $R(a)R(b)<0$ .

Then $R(c)=0$ i.e. $P(c)-Q(c)=0$ so, $P(c)=Q(c)$

Consider the function $f(x)=P(x)-Q(x)$

As both $P(x)$ and $Q(x)$ are continuous functions (since they are polynomials), so is $f(x)$ .

Note that $f(a)=-Q(a) \\ f(b)=P(b)$ by the given hypothesis.

Now, $f(a)f(b)=-P(b)Q(a)<0 \quad (\text{By the given hypothesis})$

Hence, $f(a)$ and $f(b)$ are of opposite signs . Therefore, by the intermediate value theorem we have $f(c)=0\\ \implies P(c)=Q(c) \\ \text{for some } c \text{ lying between } a \text{ and } b$

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Note:

Even if we were merely given that $P(x)\: and \: Q(x)$ are continuous with the fact that $P(b)Q(a)>0$ the conclusion still holds.