Question

If the difference of the roots of the quadratic equation is 4 and the difference of their cubes is 208, find the quadratic equation.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{be}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}\mathrm{equation}. \\ \mathrm{Then}, \\ \mathrm{\alpha }-\mathrm{\beta }=4 \\ \mathrm{and}{\mathrm{\alpha }}^3-{\mathrm{\beta }}^3=208 \\ =>(\mathrm{\alpha }-\mathrm{\beta }{)}^3+3\mathrm{\alpha \beta }(\mathrm{\alpha }-\mathrm{\beta })=208 \\ =>(4{)}^3+3\mathrm{\alpha \beta }\times 4=208 \\ =>64+12\mathrm{\alpha \beta }=208 \\ =>12\mathrm{\alpha \beta }=208-64=144 \\ =>\mathrm{\alpha \beta }=\frac{144}{12}=12 \\ \mathrm{Now},\mathrm{on}\mathrm{using}\mathrm{the}\mathrm{identity}(a\mathit{+}b{)}^2=(a\mathit{-}b{)}^2+4ab,\mathrm{we}\mathrm{get}: \\ (\mathrm{\alpha }+\mathrm{\beta }{)}^2=(\mathrm{\alpha }-\mathrm{\beta }{)}^2+4\mathrm{\alpha \beta } \\ =(4{)}^2+4\times 12 \\ =16+48 \\ =64 \\ =>\mathrm{\alpha }+\mathrm{\beta }=\pm 8 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{if}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{a}\mathrm{quadratic}\mathrm{equation},\mathrm{the}\mathrm{quadratic}\mathrm{equation}\mathrm{is} \\ {x}^2\pm 8x+12=0 \end{gathered}$$