Question

If $\alpha$ and $\beta$ are the roots of $a{x}^2+c=bx$, then the equation $(a+cy{)}^2=b^{2}y$ in $y$ has the roots

• A. ${\alpha }^{-1},{\beta }^{-1}$
• B. ${\alpha }^2,{\beta }^2$
• C. $\alpha {\beta }^{-1},{\alpha }^{-1}\beta$
• D. ${\alpha }^{-2},{\beta }^{-2}$

Answer 1

The correct option is D ${\alpha }^{-2},{\beta }^{-2}$

$\alpha$ and $\beta$ are the roots of $a{x}^2+c=bx$

i.e. $a{x}^2-bx+c=0$

$\Rightarrow \alpha +\beta =\frac{b}{a},\alpha \beta =\frac{c}{a}$

Also, $(a+cy{)}^2=b^{2}y$

$\Rightarrow c^2{y}^2+a^2+2acy=b^{2}y$

$\Rightarrow c^2{y}^2-(b^2-2ac)y+a^2=0$

$\Rightarrow {\left(\frac{c}{a}\right)}^2{y}^2-\{{\left(\frac{b}{a}\right)}^2-2\left(\frac{c}{a}\right)\}y+1=0$

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

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

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

$\Rightarrow y^2-({\alpha }^{-2}+{\beta }^{-2})y+{\alpha }^{-2}{\beta }^{-2}=0$

$\Rightarrow (y-{\alpha }^{-2})(y-{\beta }^{-2})=0$

So, the roots are ${\alpha }^{-2}$ and ${\beta }^{-2}$