Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-3x-1=0$, then form a quadratic equation whose roots are$\frac{1}{{\alpha }^2}$ and $\frac{1}{{\beta }^2}$.

• A. $x^2-11x+1=0$
• B. $x^2+11x+1=0$
• C. $x^2-11x-1=0$
• D. None of these

Answer 1

Answer: (A)

$x^2-11x+1=0$

$\Rightarrow$ $\alpha$ and $\beta$ are the roots of the equation $x^2-3x-1=0$

$\Rightarrow$ Here. $a=1,b=-3,c=-1$

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{-1}{1}=-1$ ----- ( 1 )

$\Rightarrow$ ${\alpha }^2{\beta }^2=(-1{)}^2=1$ ----- ( 2 )

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-(-3)}{1}=3$ ----- ( 3 )

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow$ $(3{)}^2={\alpha }^2+{\beta }^2+2(-1)$ [ From ( 1 ) and ( 3 ) ]

$\therefore$ ${\alpha }^2+{\beta }^2=9+2=11$ ----- ( 4 )

$\Rightarrow$ $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}=\frac{11}{1}=11$ ---- ( 5 ) [ By using ( 2 ) and ( 4 ) ]

$\Rightarrow$ $\frac{1}{{\alpha }^2}.\frac{1}{{\beta }^2}=\frac{1}{{\alpha }^2{\beta }^2}=\frac{1}{1}=1$ ---- ( 6 )

Now new equation,

$\Rightarrow$ $x^2-(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2})x+\left(\frac{1}{{\alpha }^2{\beta }^2}\right)=0$

From ( 5 ) and ( 6 ),

$\Rightarrow$ $x^2-11x+1=0$