Question

Form the quadratic equation whose one of the root is
$1-3\sqrt{5}$

Answer 1

If one root of the quadratic equation is $1-3\sqrt{5}$ , the other root is $1+3\sqrt{5}$ .

$$\begin{gathered} \mathrm{Thus},\mathrm{\alpha }=1-3\sqrt{5}\mathrm{and}\mathrm{\beta }=1+3\sqrt{5} \\ \mathrm{Now}, \\ \mathrm{\alpha }+\mathrm{\beta }=1-3\sqrt{5}+1+3\sqrt{5} \\ =2 \\ \mathrm{\alpha \beta }=(1-3\sqrt{5})(1+3\sqrt{5}) \\ =(1{)}^2-{\left(3\sqrt{5}\right)}^2 \\ =1-45 \\ =-44 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{if}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{a}\mathrm{quadratic}\mathrm{equation},\mathrm{the}\mathrm{equation}\mathrm{is} \\ {x}^2-(\mathrm{\alpha }+\mathrm{\beta })x+\mathrm{\alpha \beta }=0 \\ \mathrm{On}\mathrm{substituting}\mathrm{\alpha }+\mathrm{\beta }=2\mathrm{and}\mathrm{\alpha \beta }=-44,\mathrm{we}\mathrm{get}: \\ {x}^2-(2)x+(-44)=0 \\ =>x^2-2x-44=0 \end{gathered}$$