Question

find the zeros of the quadratic polynomial root 3 x square - 8 x + 4 root 3 also find Alpha upon beta whole square minus beta upon Alpha whole square

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}\sqrt{3}{\mathrm{x}}^2-8\mathrm{x}+4\sqrt{3}=0 \\ \mathrm{Now},\mathrm{\alpha }+\mathrm{\beta }=\frac{-\mathrm{coeff}.\mathrm{of}\mathrm{x}}{\mathrm{coeff}.\mathrm{of}{\mathrm{x}}^2} \\ =\frac{-\left(-8\right)}{\sqrt{3}} \\ =\frac{8}{\sqrt{3}}...\left(1\right) \\ \mathrm{and}\mathrm{\alpha \beta }=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coeff}.\mathrm{of}{\mathrm{x}}^2} \\ =\frac{4\sqrt{3}}{\sqrt{3}} \\ =4 \\ \mathrm{Now}\mathrm{we}\mathrm{know}\mathrm{that}{\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2={\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-4\mathrm{\alpha \beta } \\ \Rightarrow {\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2={\left(\frac{8}{\sqrt{3}}\right)}^2-4\left(4\right) \\ \Rightarrow {\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2=\frac{64}{3}-16 \\ \Rightarrow {\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2=\frac{64-48}{3}=\frac{16}{3} \\ \Rightarrow \mathrm{\alpha }-\mathrm{\beta }=\frac{\pm 4}{\sqrt{3}}...\left(2\right) \\ \mathrm{adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2\mathrm{\alpha }=\frac{8}{\sqrt{3}}\pm \frac{4}{\sqrt{3}} \\ \mathrm{Case}1.2\mathrm{\alpha }=\frac{8}{\sqrt{3}}+\frac{4}{\sqrt{3}}=\frac{12}{\sqrt{3}} \\ 2\mathrm{\alpha }=\frac{12}{\sqrt{3}} \\ \mathrm{\alpha }=\frac{6}{\sqrt{3}}=\frac{6\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=2\sqrt{3} \\ \mathrm{Case}2.2\mathrm{\alpha }=\frac{8}{\sqrt{3}}-\frac{4}{\sqrt{3}} \\ 2\mathrm{\alpha }=\frac{2}{\sqrt{3}} \\ \mathrm{\alpha }=\frac{1}{\sqrt{3}} \\ \mathrm{When}\mathrm{\alpha }=2\sqrt{3} \\ \mathrm{So},\mathrm{\alpha \beta }=4 \\ \Rightarrow \mathrm{\beta }=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3} \\ \mathrm{Consider}{\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}\right)}^2-{\left(\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)}^2 \\ ={\left(\frac{2\sqrt{3}}{\frac{2\sqrt{3}}{3}}\right)}^2-{\left(\frac{\frac{2\sqrt{3}}{3}}{2\sqrt{3}}\right)}^2 \\ ={\left(\frac{6\sqrt{3}}{2\sqrt{3}}\right)}^2-{\left(\frac{2\sqrt{3}}{3\times 2\sqrt{3}}\right)}^2 \\ =3^2-\frac{1}{3^2} \\ =9-\frac{1}{9} \\ =\frac{81-1}{9} \\ =\frac{80}{9} \\ \mathrm{Similarly}\mathrm{try}\mathrm{with}\mathrm{\alpha }=\frac{1}{\sqrt{3}}\mathrm{and}\mathrm{\beta }=4\sqrt{3} \end{gathered}$$
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