Let f:[0,5] ->[0,5] be an invertible function defined by f(x) = ax^2 + bx + c, where abc is not equal to zero, then the roots of equation cx^2 + bx + a are:

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ \mathrm{For} \mathrm{roots} \\ {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0 \\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ \mathrm{Now} \mathrm{replace} \mathrm{x} \mathrm{with} \frac{1}{\mathrm{x}} \\ \frac{\mathrm{a}}{{\mathrm{x}}^2}+\frac{\mathrm{b}}{\mathrm{x}}+\mathrm{c}=0 \\ \Rightarrow {\mathrm{cx}}^2+\mathrm{bx}+\mathrm{a}=0 \\ \mathrm{Hence} \mathrm{roots} \mathrm{of} {\mathrm{cx}}^2+\mathrm{bx}+\mathrm{a}=0 \mathrm{will} \mathrm{be} \mathrm{reciprocal} \mathrm{of} \mathrm{roots} \mathrm{of} \\ {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0 \\ \mathrm{Hence} \mathrm{roots} \mathrm{will} \mathrm{be} \frac{2\mathrm{a}}{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}} \end{gathered}$$

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