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Question

If the polynomial equation ${a}_{0}{x}^{n}+{a}_{n-1}{x}^{n-1}+{a}_{n-2}{x}^{n-2}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}=0$ n positive integer, has two different real roots α and β, then between α and β, the equation $n{a}_{n}{x}^{n-1}+\left(n-1\right){a}_{n-1}{x}^{n-2}+...+{a}_{1}=0\mathrm{has}$ (a) exactly one root (b) almost one root (c) at least one root (d) no root

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Solution

(c) at least one root We observe that, $n{a}_{n}{x}^{n-1}+\left(n-1\right){a}_{n-1}{x}^{n-2}+...+{a}_{1}=0$ is the derivative of the polynomial ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+{a}_{n-2}{x}^{n-2}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}=0$ Polynomial function is continuous every where in R and consequently derivative in R Therefore, ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+{a}_{n-2}{x}^{n-2}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$ is continuous on $\left[\alpha ,\beta \right]$ and derivative on $\left(\alpha ,\beta \right)$. Hence, it satisfies the both the conditions of Rolle's theorem. By algebraic interpretation of Rolle's theorem, we know that between any two roots of a function $f\left(x\right)$, there exists at least one root of its derivative. Hence, the equation $n{a}_{n}{x}^{n-1}+\left(n-1\right){a}_{n-1}{x}^{n-2}+...+{a}_{1}=0$ will have at least one root between $\alpha \mathrm{and}\beta$.

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