Question

If $\alpha$ and $\beta$ are non-zero distinct complex numbers satisfying ${\alpha }^2+3=5\alpha$ and ${\beta }^2=5\beta -3$ then a quadratic equation having $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as its roots is

• A. $3x^2-37x+3=0$
• B. $3x^2-31x+3=0$
• C. $3x^2-19x+3=0$
• D. $3x^2-17x+3=0$

Answer 1

Answer: (C)

Given, ${\alpha }^2=5\alpha -3\Rightarrow {\alpha }^2-5\alpha +3=0$
$\text{ and }{\beta }^2=5\beta -3\Rightarrow {\beta }^2-5\beta +3=0$

Now, $\alpha +\beta =5$ and $\alpha \beta =3$
$\therefore \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{(5{)}^2-2\times 3}{3}=\frac{19}{3}=$ sum of roots.
Also, $\left(\frac{\alpha }{\beta }\right)\left(\frac{\beta }{\alpha }\right)=1=$ product of roots.
$\therefore$ The quadratic equation having $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as its roots is
$x^2-\frac{19}{3}x+1=0\Rightarrow 3x^2-19x+3=0$