Question

Let $\alpha \ne \beta$, ${\alpha }^2+3=5\alpha$ and ${\beta }^2=5\beta -3$. The quadratic equation whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ will be:

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

Answer 1

The correct option is A

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

We have,
${\alpha }^2-5\alpha +3=0\left(1\right)$
${\beta }^2-5\beta +3=0\left(2\right)$

Subtracting (2) from (1), we get
${\alpha }^2-5\alpha +3-{\beta }^2+5\beta +3=0$
$\Rightarrow {\alpha }^2-{\beta }^2-5(\alpha -\beta )=0$
$\Rightarrow (\alpha -\beta )\left(\right(\alpha +\beta -5\left)\right)=0$
But $\alpha \ne \beta \therefore \alpha +\beta =5\left(3\right)$

Adding (1) and (2), we get
$\Rightarrow {\alpha }^2+{\beta }^2-5(\alpha +\beta )+6=0$
$\Rightarrow {\alpha }^2+{\beta }^2-5\left(5\right)+6=0\left[From\right(3\left)\right]$
$\Rightarrow {\alpha }^2+{\beta }^2=19\left(4\right)$

Now, $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$
$\Rightarrow 25=19+2\alpha \beta$
$\Rightarrow 2\alpha \beta =6\Rightarrow \alpha \beta =3\left(5\right)$

The roots of the quadratic equation are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$
$\therefore Sumofroots=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{19}{3}$
$andproductofroots=\frac{\alpha }{\beta }\times \frac{\beta }{\alpha }=1$

$\therefore$ Required quadratic equation is
$x^2-\frac{19}{3}x+1=0$
$\Rightarrow 3x^2-19x+3=0$