Question

If $\alpha \ne \beta$ and ${\alpha }^2=5\alpha -3,{\beta }^2=5\beta -3,$ then the equation having $\alpha /\beta$ and $\beta /\alpha$ as its roots, is :

• A. $3x^2+19x+3=0$
• B. $3x^2-19x+3=0$
• C. $3x^2-19x-3=0$
• D. $x^2-16x+1=0$

Answer 1

Answer: (B)

Key Idea : The equation having $\alpha$ and $\beta$ as its
roots, is $x^2-(\alpha +\beta )x+\alpha \beta =0$
Since, ${\alpha }^2=5\alpha -3$
$\Rightarrow {\alpha }^2-5\alpha +3=0$
and ${\beta }^2=5\beta -3$
$\Rightarrow {\beta }^2-5\beta +3=0$
These two equations shows that $\alpha$ and $\beta$ are the roots of the equation
$x^2-5x+3=0$
$\therefore \alpha +\beta =5$ and $\alpha \beta =3$
Now
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$
$=\frac{25-6}{3}=\frac{19}{3}$
and $\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=1$
Thus the equation having $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as its roots is given by
$x^2-\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\right)x+\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=0$
$\Rightarrow x^2-\frac{19}{3}x+1=0$
$\Rightarrow 3x^2-19x+3=0$