Question

if alpha and beta are the roots of the equation x²-3x+4=0 then value of alpha³-4alpha²+7alpha-2 is
options are
a)0 b)1 c)2

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}:\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{are}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2-3\mathrm{x}+4=0 \\ \mathrm{\alpha }+\mathrm{\beta }=\frac{-\mathrm{coeff}.\mathrm{of}\mathrm{x}}{\mathrm{coeff}.\mathrm{of}{\mathrm{x}}^2}=\frac{-\left(-3\right)}{1}=3 \\ \Rightarrow \mathrm{\beta }=3-\mathrm{\alpha }....\left(1\right) \\ \mathrm{\alpha \beta }=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coeff}.\mathrm{of}{\mathrm{x}}^2}=\frac{4}{1}=4 \\ \Rightarrow \mathrm{\beta }=\frac{4}{\mathrm{\alpha }}...\left(2\right) \\ \mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 3-\mathrm{\alpha }=\frac{4}{\mathrm{\alpha }} \\ \Rightarrow -{\mathrm{\alpha }}^2+3\mathrm{\alpha }-4=0 \\ \Rightarrow {\mathrm{\alpha }}^2-3\mathrm{\alpha }+4=0 \\ \Rightarrow \mathrm{\alpha }=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\left[\mathrm{using}\mathrm{quadratic}\mathrm{formula}\right] \\ \Rightarrow \mathrm{\alpha }=\frac{3\pm \sqrt{9-16}}{2} \\ \Rightarrow \mathrm{\alpha }=\frac{3\pm \mathrm{i}\sqrt{7}}{2} \\ \mathrm{Case}\mathrm{I}:\mathrm{When}\mathrm{\alpha }=\frac{3+\mathrm{i}\sqrt{7}}{2} \\ \mathrm{Consider}{\mathrm{\alpha }}^3-4{\mathrm{\alpha }}^2+7\mathrm{\alpha }-2 \\ ={\left(\frac{3+\mathrm{i}\sqrt{7}}{2}\right)}^3-4{\left(\frac{3+\mathrm{i}\sqrt{7}}{2}\right)}^2+7\left(\frac{3+\mathrm{i}\sqrt{7}}{2}\right)-2 \\ =\frac{-36+\mathrm{i}20\sqrt{7}}{8}-\left(2+\mathrm{i}6\sqrt{7}\right)+\frac{21+\mathrm{i}7\sqrt{7}}{2}-2 \\ =\frac{-36+\mathrm{i}20\sqrt{7}-8\left(2+\mathrm{i}6\sqrt{7}\right)+4\left(21+\mathrm{i}7\sqrt{7}\right)-16}{8} \\ =\frac{-36+\mathrm{i}20\sqrt{7}-16-\mathrm{i}48\sqrt{7}+84+\mathrm{i}28\sqrt{7}-16}{8} \\ =\frac{16+0}{8} \\ =2 \\ \mathrm{Similarly}\mathrm{try}\mathrm{with}\mathrm{\alpha }=\frac{3-\mathrm{i}\sqrt{7}}{2} \\ \mathrm{Note}: \\ {\left(\mathrm{a}+\mathrm{b}\right)}^3={\mathrm{a}}^3+{\mathrm{b}}^3+3{\mathrm{a}}^2+3{\mathrm{ab}}^2 \end{gathered}$$
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