Question

If $\alpha$ and $\beta$ are roots of equation $x^3-2x+3=0$,then the equation whose roots are $\frac{\alpha -1}{\alpha +1}$ and $\frac{\beta -1}{\beta +1}$ will be

• A. $3x^2-x-1=0$
• B. $3x^2+2x+1=0$
• C. $3x^2-x+1=0$
• D. $x^2-2x+1=0$

Answer 1

The correct option is C $3x^2-x+1=0$

$x^3-2x+3=0$

$\alpha +\beta =2$

$\alpha \beta =3$

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

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

$\Rightarrow x^2(\alpha \beta +(\alpha +\beta )+1)-x(\alpha \beta +\alpha -\beta -1+\alpha \beta +\beta -\alpha -1)+(\alpha \beta -(\alpha +\beta )+1)=0$

$\Rightarrow x^2(3+2+1)-x(3-1)+(3-2+1)=0$

$\Rightarrow 6x^2-2x+2=0$

$\Rightarrow 3x^2-x+1=0$