Question

The quadratic polynomial whose zeros are $3+\sqrt{5}$ and $3-\sqrt{5}$ is

• A. $y^2-6y+4$
• B. $2y^2-6y+4$
• C. $2y^2-12y+8$
• D. $y^2+10y+4$

Answer 1

The correct option is A $y^2-6y+4$

$\Rightarrow$ $3+\sqrt{5}$ and $3-\sqrt{5}$ are roots of the required equation.

$\Rightarrow$ Let $\alpha =3+\sqrt{5}$ and $\beta =3-\sqrt{5}$

$\Rightarrow$ $\alpha +\beta =3+\sqrt{5}+3-\sqrt{5}=6$ ------ ( 1 )

$\Rightarrow$ $\alpha \beta =(3+\sqrt{5})(3-\sqrt{5})=9-3\sqrt{5}+3\sqrt{5}-5=4$ ---- ( 2 )

Required equation,

$\Rightarrow$ $y^2-(\alpha +\beta )y+\left(\alpha \beta \right)=0$

Using ( 1 ) and ( 2 ),

$\therefore$ $y^2-6y+4=0$