Question

If $\alpha \ne \beta ,{\alpha }^2=5\alpha -3$, and ${\beta }^2=5\beta -3$, 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. $x^2+5x+3=0$

Answer 1

The correct option is A $3x^2-19x+3=0$

${\alpha }^2=5\alpha -3{\beta }^2=5\beta -3$

Equation is $x^2-5x+3=0$

$\alpha +\beta =5\alpha \beta =3$

If $\frac{\alpha }{\beta },\frac{\beta }{\alpha }$ are roots

Sum of roots$=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{25-2\left(3\right)}{3}=\frac{19}{3}$

Products of roots$=\frac{\alpha }{\beta }\times \frac{\beta }{\alpha }=1$

$x^2-\frac{19}{3}x+1=0$

$\therefore 3x^2-19x+3=0$