Question

find a quadratic polynomial with zeroes 3 + root 2 and 3 - root 2

Answer 1

$$\begin{gathered} \mathrm{Let}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{zeroes}\mathrm{of}\mathrm{required}\mathrm{quadratic}\mathrm{polynomial}p\left(x\right). \\ \mathrm{So},\alpha =3+\sqrt{2};\beta =3-\sqrt{2} \\ \mathrm{Now},\alpha +\beta =\left(3+\sqrt{2}\right)+\left(3-\sqrt{2}\right)=6 \\ \alpha \beta =\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)={\left(3\right)}^2-{\left(\sqrt{2}\right)}^2=9-2=7 \\ \mathrm{Now},\mathrm{required}\mathrm{quadratic}\mathrm{polynomial}\mathrm{is}, \\ p\left(x\right)=k\left[x^2-\left(\alpha +\beta \right)x+\alpha \beta \right],k\mathrm{is}\mathrm{non}-\mathrm{zero}\mathrm{real}\mathrm{number} \\ \Rightarrow p\left(x\right)=k\left[x^2-6x+7\right] \\ \mathrm{Taking}k=1,\mathrm{we}\mathrm{get}, \\ p\left(x\right)=x^2-6x+7 \end{gathered}$$