Question

If the quadratic equation 2x² – (a³ + 8a – 1) x + a² – 4a = 0 possesses roots of opposite signs, then a lies in the interval ____________.

Answer 1

$$\begin{gathered} \mathrm{For}2x^2–\left(a^3+8a–1\right)x+a^2–4a=0\mathrm{has}\mathrm{roots}\mathrm{of}\mathrm{opposite}\mathrm{sign}. \\ \mathrm{Let}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{be}2\mathrm{roots}\mathrm{of}\mathrm{above}\mathrm{equation}. \\ \mathrm{Then},\mathrm{according}\mathrm{to}\mathrm{give}\mathrm{condition}, \\ \alpha \beta <0 \\ \mathrm{also},\mathrm{since}\mathrm{sum}\mathrm{of}\mathrm{roots}=\frac{a^3+8a-1}{2} \\ \Rightarrow \alpha +\beta =\frac{a^3+8a-1}{2} \\ \mathrm{and}\mathrm{product}\mathrm{of}\mathrm{roots}\mathrm{is}{a}^2-4a \\ \mathrm{i}.\mathrm{e}.\alpha \beta =a^2-4a \\ \mathrm{since}\alpha \beta <0 \\ \Rightarrow a^2-4a<0 \\ \mathrm{i}.\mathrm{e}.a\left(a-4\right)<0 \end{gathered}$$

Since a(a – 4) < 0 is true for 0 < a < 4
Hence a ∈ (0, 4)