Question

Form the quadratic equation, whose roots are $3+\sqrt{2}$ and $3-\sqrt{2}$.

Answer 1

Let $x=3+\sqrt{2}orx=3-\sqrt{2}$ are the roots of a quadratic equation.

$\therefore$ $x-(3+\sqrt{2})=0orx-(3-\sqrt{2})=0$

$⟹$ $(x-(3+\sqrt{2}\left)\right)(x-(3-\sqrt{2}\left)\right)=0$

$⟹$ $(x^2-x(3-\sqrt{2})-x(3+\sqrt{2})+(3+\sqrt{2}\left)\right(3-\sqrt{2}\left)\right)=0$

$⟹$ $(x^2-3x+x\sqrt{2}-3x-x\sqrt{2}+9-2)=0$

$⟹$ $(x^2-6x+7)=0$

Hence $(x^2-6x+7)=0$ is the quadratic equation having roots $3+\sqrt{2}$ and $3-\sqrt{2}$.