Question

Find the roots $\alpha ,\beta ,\gamma$ of $x^3-11x^2+36x-36=0$ if $\frac{2}{\beta }=\frac{1}{\alpha }+\frac{1}{\gamma }$

Answer 1

Given that,

$\alpha ,\beta ,\gamma$ are the roots of $x^3-11x^2+36x-36=0$

Here, $a=1,b=-11,c=36,d=-36$

$\alpha +\beta +\gamma =\frac{-b}{a}=11$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=36$

$\alpha \beta \gamma =\frac{-d}{a}=36$

$\because \frac{2}{\beta }=\frac{1}{\alpha }+\frac{1}{\gamma }$

$⟹\frac{2}{\beta }=\frac{\alpha +\gamma }{\alpha \gamma }$

$⟹2\alpha \gamma =\alpha \beta +\beta \gamma$

$⟹3\alpha \gamma =\alpha \beta +\beta \gamma +\alpha \gamma =36$

$⟹\alpha \gamma =\frac{36}{3}=12$....................(i)

Now, $\beta =\frac{\alpha \beta \gamma }{\alpha \gamma }=\frac{36}{12}=3$

$⟹\alpha +\gamma =11-3=8$................................(ii)

From (i) & (ii) ,

$\alpha ,\beta$ are $6$ and $2$

Hence, $\alpha =6$

$\beta =3$

$\gamma =2$