Question

Form the quadratic equation if its one of the root is

$3-2\sqrt{5}$

Answer 1

If one root of the quadratic equation is $3-2\sqrt{5}$ , the other root of the equation is $3+2\sqrt{5}$.
Thus, $\alpha =3-2\sqrt{5}$ and $\beta =3+2\sqrt{5}$
Now,
$$\begin{gathered} \alpha +\beta =3-2\sqrt{5}+3+2\sqrt{5}=6 \\ \mathrm{and} \\ \alpha \beta =\left(3-2\sqrt{5}\right)\left(3+2\sqrt{5}\right)={\left(3\right)}^2-{\left(2\sqrt{5}\right)}^2=9-20=-11 \end{gathered}$$

We know that if α and β are the roots of a quadratic equation, then the quadratic equation is
x² – (α + β)x + α β = 0
On substituting α + β = 6 and α β = –11, we get:
x² – (6)x + (–11) = 0
$\Rightarrow$ x² – 6x – 11 = 0