Question

If $\alpha \neq \beta$ 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. $3 x^{2}+19 x+3=0$
• B. $3 x^{2}-19 x+3=0$
• C. $3 x^{2}-19 x-3=0$
• D. $x^{2}-16 x+1=0$

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
$x^{2}-5 x+3=0$
$\therefore \alpha+\beta=5$ 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 $
$\Rightarrow x^{2}-\frac{19}{3} x+1=0$
$\Rightarrow 3 x^{2}-19 x+3=0$