Question

If $\alpha \ne \beta$ and ${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$, find the equation whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$.

• A. $x^2-15x+30=0$
• B. $x^2-25x+29=0$
• C. $3x^2-19x+3=0$
• D. $(3x-1)(3x-2)=0$

Answer 1

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

As,

${\alpha }^2=5\alpha -3\Rightarrow {\alpha }^2-5\alpha +3=0$

${\beta }^2=5\beta -3\Rightarrow {\beta }^2-5\beta +3=0$

Therefore, $\alpha ,\beta$ are the roots of the equation
$x^2=5x-3$
$x^2-5x+3=0$

Now,
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$
$=\frac{{\alpha }^2+{\beta }^2}{\alpha .\beta }$
$=\frac{(\alpha +\beta {)}^2-2\alpha .\beta }{\alpha .\beta }$
$=\frac{25-6}{3}$
$=\frac{19}{3}$

Hence,
The required equation is
$x^2-\frac{19}{3}x+1=0$
$3x^2-19x+3=0$.