Question

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

• A. $3x^2-2x+1=0$
• B. $x^2-x+3=0$
• C. $5x^2-2x+3=0$
• D. $x^2-2x+7=0$

Answer 1

Answer: (A)

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

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

$\Rightarrow$ Here, $a=1,b=-2,c=3$

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-(-2)}{1}=2$ ---- ( 1 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{3}{1}=3$ ----- ( 2 )

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

$=\frac{\alpha \beta +\alpha -\beta -1+\alpha \beta +\beta -\alpha -1}{\alpha \beta +\alpha +\beta +1}$

$=\frac{2\alpha \beta -2}{\alpha \beta +\alpha +\beta +1}$

$=\frac{2\left(3\right)-2}{3+2+1}$ [ Using ( 1 ) and ( 2 )]

$=\frac{4}{6}$

$\therefore$ $\frac{\alpha -1}{\alpha +1}+\frac{\beta -1}{\beta +1}=\frac{2}{3}$ ---- ( 3 )

$\Rightarrow$ $\frac{\alpha -1}{\alpha +1}.\frac{\beta -1}{\beta +1}=\frac{\alpha \beta -\alpha -\beta +1}{\alpha \beta +\alpha +\beta +1}$

$=\frac{\alpha \beta -(\alpha +\beta )+1}{\alpha \beta +\alpha +\beta +1}$

$=\frac{3-2+1}{3+2+1}$ [ From ( 1 ) and ( 2 )]

$\therefore$ $\frac{\alpha -1}{\alpha +1}.\frac{\beta -1}{\beta +1}=\frac{1}{3}$ ----- ( 4 )

Now new equation,

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

Now using ( 3 ) and ( 4 ),

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

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