Question

The quadratic equation whose roots are $\alpha$ and $\beta$ is given by

• A. $x^2-2x-1=0$
• B. $x^2-2x-2=0$
• C. $x^2-x-1=0$
• D. None

Answer 1

The correct option is C $x^2-x-1=0$

Given that, $T_n=A{\alpha }^{n-1}+B{\beta }^{n-1}$
$\because T_1=0$
$\therefore A+B=0$
$\Rightarrow A=-B$ ....(1)
$\because T_2=1$
$\therefore A\alpha +B\beta =1$
$\Rightarrow A(\alpha -\beta )=1$ ....(2) [ From (1) ]
$\because T_3=1$
$\therefore A{\alpha }^2+B{\beta }^2=1$
$\Rightarrow A({\alpha }^2-{\beta }^2)=1$ ....[ From (1) ]
$\Rightarrow A(\alpha -\beta )(\alpha +\beta )=1$
$\Rightarrow \alpha +\beta =1$ ....(3) [ From (2) ]
$\because T_4=2$
$\therefore A{\alpha }^3+B{\beta }^3=2$
$\Rightarrow A({\alpha }^3-{\beta }^3)=2$ ....[ From (1) ]

$\Rightarrow A(\alpha -\beta )({\alpha }^2+{\beta }^2+\alpha \beta )=2$
$\Rightarrow {\alpha }^2+{\beta }^2+\alpha \beta =2$ ....(4) [ From (2) ]
$\Rightarrow \alpha \beta ={(\alpha +\beta )}^2-1$
$\Rightarrow \alpha \beta =-1$ ....(5) [ From (3) ]
If $\alpha$ and $\beta$ are roots of a quadratic equation, then
$x^2-(\alpha +\beta )x+\alpha \beta =0$
$\Rightarrow x^2-x-1=0$ ....[ From (3) & (5) ]
Ans: C