Question

alpha and beta are the roots of 3x^2+5x-7 then the value of {1/3alpha+5}^2+{1/3beta+5}^2

Answer 1

$$\begin{gathered} \mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{of}3{\mathrm{x}}^2+5\mathrm{x}-7=0 \\ \mathrm{\alpha }+\mathrm{\beta }=\frac{-5}{3} \\ \mathrm{\alpha \beta }=\frac{-7}{3} \\ \mathrm{Also},3{\mathrm{\alpha }}^2+5\mathrm{\alpha }-7=0 \\ \Rightarrow 9{\mathrm{\alpha }}^2+15\mathrm{\alpha }-21=0 \\ \Rightarrow (3\mathrm{\alpha }+5{)}^2-15\mathrm{\alpha }-21-25=0 \\ \Rightarrow (3\mathrm{\alpha }+5{)}^2=15\mathrm{\alpha }+46 \\ \mathrm{Similarly},(3\mathrm{\beta }+5{)}^2=15\mathrm{\beta }+46 \\ \mathrm{Now},\frac{1}{(3\mathrm{\alpha }+5{)}^2}+\frac{1}{(3\mathrm{\beta }+5{)}^2} \\ =\frac{1}{15\mathrm{\alpha }+46}+\frac{1}{15\mathrm{\beta }+46} \\ =\frac{15(\mathrm{\alpha }+\mathrm{\beta })+92}{(15\mathrm{\alpha }+46)(15\mathrm{\beta }+46)}=\frac{15(\mathrm{\alpha }+\mathrm{\beta })+92}{{15}^2\mathrm{\alpha \beta }+15\times 46(\mathrm{\alpha }+\mathrm{\beta })+{46}^2} \\ =\frac{15\left({\displaystyle \frac{-5}{3}}\right)+92}{{15}^2\left({\displaystyle \frac{-7}{3}}\right)+15\times 46\left({\displaystyle \frac{-5}{3}}\right)+{46}^2}=\frac{67}{441} \end{gathered}$$