Question

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

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

Answer 1

Answer: (C)

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

Let $\alpha ,\beta$ are roots of $x^2-2x+3=0$
Then $\alpha +\beta =2;\alpha \beta =3$.
Now $\frac{\alpha -1}{\alpha +1}+\frac{\beta -1}{\beta +1}=\frac{(\alpha -1)(\beta +1)+(\alpha +1)(\beta -1)}{(\alpha +1)(\beta +1)}$
$=\frac{\alpha \beta +\alpha -\beta -1+\alpha \beta -\alpha +\beta -1}{\alpha \beta +\alpha +\beta +1}$
$=\frac{2\alpha \beta -2}{\alpha \beta +\alpha +\beta +1}=\frac{6-2}{3+2+1}=\frac{4}{6}=\frac{2}{3}$
$\left(\frac{\alpha -1}{\alpha +1}\right)\left(\frac{\beta -1}{\beta +1}\right)=\frac{\alpha \beta -\alpha -\beta +1}{(\alpha \beta +\alpha +\beta +1)}$
$=\frac{3-2+1}{3+2+1}=\frac{2}{6}=\frac{1}{3}$
Hence, equation is $x^2-\left(\frac{2}{3}\right)x+\left(\frac{1}{3}\right)=0\Rightarrow {3x}^2-2x+1=0$.