Question

If the difference of the roots of the quadratic equation is 3 and difference between the cubes is 189. find the quadratic equation.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{be}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}\mathrm{equation}.\mathrm{Then}, \\ \mathrm{\alpha }-\mathrm{\beta }=3\mathrm{and} \\ {\mathrm{\alpha }}^3-{\mathrm{\beta }}^3=189 \\ =>(\mathrm{\alpha }-\mathrm{\beta }{)}^3+3\mathrm{\alpha \beta }(\mathrm{\alpha }-\mathrm{\beta })=189 \\ =>(3{)}^3+3\mathrm{\alpha \beta }\times 3=189 \\ =>27+9\mathrm{\alpha \beta }=189 \\ =>9\mathrm{\alpha \beta }=189-27=162 \\ =>\mathrm{\alpha \beta }=\frac{162}{9}=18 \\ \mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{identity}(a\mathit{+}b{)}^2=(a\mathit{-}b{)}^2+4ab,\mathrm{we}\mathrm{get}: \\ (\mathrm{\alpha }+\mathrm{\beta }{)}^2=(\mathrm{\alpha }-\mathrm{\beta }{)}^2+4\mathrm{\alpha \beta } \\ =(3{)}^2+4\times 18 \\ =9+72 \\ =81 \\ =>\mathrm{\alpha }+\mathrm{\beta }=\pm 9 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{if}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{a}\mathrm{quadratic}\mathrm{equation},\mathrm{the}\mathrm{quadratic}\mathrm{equation}\mathrm{is} \\ {x}^2-(\mathrm{\alpha }+\mathrm{\beta })x+\mathrm{\alpha \beta }=0 \\ \mathrm{On}\mathrm{subtituting}\mathrm{\alpha }+\mathrm{\beta }=\pm 9\mathrm{and}\mathrm{\alpha \beta }=18,\mathrm{we}\mathrm{get}: \\ {x}^2\pm 9x+18=0 \end{gathered}$$