Let $V_n$ be the vector space generated by the set $\{1, x, \ldots, x^n\}$.
We say that two polynomials $f(x)$ and $g(x)$ are orthogonal (with respect to a inner product) if
$$\langle f(x), g(x) \rangle = 0.$$
Consider the inner product
$$\langle f(x), g(x) \rangle = \int_{a}^{b}f(x)g(x)dx.$$
Let $\{p_0(x), \ldots, p_n(x)\}$ and $\{q_0(x), \ldots, q_n(x)\}$ two orthogonal basis of $V_n$ with respect to the inner product above such that the $p_j(x)$ and $q_j(x)$ have degree $j$, $j \in \{0, \ldots, n\}$. It's not hard to prove that $p_n(x)$ is orthogonal to all polynomials with degree less than $n$.
The following preposition is what I mention in the comments (and I know how to prove).
Preposition 1. The polynomial $p_n(x)$ has $n$ distinct roots in $(a,b)$ for any $n \geq 1$.
The other preposition (that I'm trying to prove) is the following:
Preposition 2. The polynomials $p_n$ and $q_n$ have the same roots. That's, if the roots of $p_n(x)$ are $x_1, \ldots, x_n$ so the roots of $q_n(x)$ also are $x_1, \ldots, x_n$.
My idea was have some contradiction with the sign of the inner product, but I don't know how to progress or if It's not the right idea:
Exists real coefficients $c_0, c_1, \ldots, c_n$ such that $q_n(x) = c_np_n(x) + \ldots + c_1p_1(x) + c_0p_0(x)$. We know that $p_n(x)$ is orthogonal to all polynomials with degree less than $n$, so
\begin{align*} \langle p_n(x), q_n(x) \rangle &= \int_{a}^{b} p_n(x)(c_n p_n(x) + \ldots + c_0p_0(x))dx \\ &= c_n\int_{a}^{b} p_n(x)p_n(x)dx \\ &= c_n\langle p_n(x),p_n(x)\rangle \end{align*}
With little loss we can assume the $p_k$ have been arranged so that they have unit norm, namely $\langle p_k, p_k \rangle = 1$.
You have defined $c_n$ such that $q_n = \sum_k c_k p_k$. Then for $k < n$, using orthogonality of the $p_k$, $$\langle q_n, p_k\rangle = c_k.$$ But the left side is also zero since $q_n$ is orthogonal to every polynomial of lower degree. Thus $c_0, c_1, c_{n-1} = 0$ and $q_n$ is a multiple of $p_n$ and both must have the same roots.