Question

If $\alpha ,\beta ,\gamma ,\delta$ are the roots of the equation $x^4+q{x}^2+rx+s=0$ find the equation whose roots are $\beta +\gamma +\delta +(\beta \gamma \delta {)}^{-1}$, and $c$.

Answer 1

Given equation, $x^4+q{x}^2+rx+s=0$ which has the roots $\alpha ,\beta ,\gamma ,\delta$

Therefore, $\sum \alpha =0,\sum \alpha \beta =q,\sum \alpha \beta \gamma =-r,\alpha \beta \gamma \delta =s$

Now, $\beta +\gamma +\delta +\frac{1}{\beta \gamma \delta }=(\alpha +\beta +\gamma +\delta )+\frac{\alpha }{\alpha \beta \gamma \delta }-\alpha =\left(\frac{1-s}{s}\right)\alpha$

Therefore, roots of the required equation are $\left(\frac{1-s}{s}\right)\alpha ,\left(\frac{1-s}{s}\right)\beta ,\left(\frac{1-s}{s}\right)\gamma ,\left(\frac{1-s}{s}\right)\delta ,c$

Or Roots of the required equation are $\lambda \alpha ,\lambda \beta ,\lambda \gamma ,\lambda \delta ,c$ where $\lambda =\frac{1-s}{s}$

For the required equation,

$S_1=\lambda \alpha +\lambda \beta +\lambda \gamma +\lambda \delta +c=\lambda (\alpha +\beta +\gamma +\delta +c)=c$

$S_2={\lambda }^2\sum \alpha \beta +\lambda (\sum \alpha )c={\lambda }^{2}q$

$S_3={\lambda }^3\sum \alpha \beta \gamma +{\lambda }^2(\sum \alpha \beta )c=-{\lambda }^{3}r+{\lambda }^{2}rc$

$S_4={\lambda }^4\alpha \beta \gamma \delta +{\lambda }^3(\sum \alpha \beta \gamma )c={\lambda }^{4}s-{\lambda }^{3}rc$

$S_5={\lambda }^4\alpha \beta \gamma \delta c={\lambda }^{4}sc$

Required equation is $x^5–S_1{x}^4+S_2{x}^3-S_3{x}^2+S_{4}x-S_5=0$

Or $x^5-c{x}^4+{\lambda }^{2}q{x}^3-(-{\lambda }^{3}r+{\lambda }^{2}rc)x^2+({\lambda }^{4}s-{\lambda }^{3}rc)x-{\lambda }^{4}sc=0$

Or $s^3{x}^5-s^{3}c{x}^4+(1-s{)}^{2}sq{x}^3+(1-s{)}^2(1-s-sc)r{x}^2+(1-s{)}^3(1-s-rc)x-(1-s{)}^{3}c=0$