Question

In the equation $x^4-p{x}^3+q{x}^2-rx+s=0$, prove that if the sum of two of the roots is equal to the sum of the other two $p^3-4pq+8r=0$; and that if the product of two of the roots is equal to the product of the other two $r^2=p^{2}s$.

Answer 1

Given, $x^4-p{x}^3+q{x}^4-rx+s=0$ and let the roots be $\alpha ,\beta ,\gamma ,\delta$

Case 1: Sum of two roots is equal to sum of other two, therefore $\alpha +\delta =\beta +\gamma \dots \dots \dots \dots \dots .\left(1\right)$

$\alpha +\beta +\gamma +\delta =p$

$⟹\alpha +\delta =\beta +\gamma =\frac{p}{2}\text{[from (1)]}\dots \dots \dots \dots \left(2\right)$

$\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta ++\gamma \delta =q$

$⟹(\alpha +\delta )(\beta +\gamma )+\alpha \delta +\beta \gamma =q$

$⟹\left(\frac{p}{2}\right)\left(\frac{p}{2}\right)+\alpha \delta +\beta \gamma =q\text{[Substituting from (2)]}$

$⟹\alpha \delta +\beta \gamma =q-\frac{1}{4}{p}^2\dots \dots \dots \dots .\left(3\right)$

$\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =r$

$⟹\beta \gamma (\alpha +\delta )+\alpha \delta (\beta +\gamma )=r$

$⟹\beta \gamma \left(\frac{p}{2}\right)+\alpha \delta \left(\frac{p}{2}\right)=r\text{[Substituting from (2)]}$

$⟹\frac{1}{2}p(\beta \gamma +\alpha \delta )=r\dots \dots \dots \dots ..\left(4\right)$

From (3) and (4), we have

$\frac{1}{2}p(q-\frac{1}{4}{p}^2)=r$

$⟹p^3-4pq+8r=0$

Case 2: Product of two roots is equal to the product of other two, therefore $\alpha \delta =\beta \gamma \dots \dots \dots \dots \dots \left(6\right)$

$\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =r$

$⟹\beta \gamma (\alpha +\delta )+\alpha \delta (\beta +\gamma )=r$

$⟹\alpha \delta (\alpha +\beta +\gamma +\delta )=r[\because \alpha \delta =\beta \gamma \text{from (6)}]$

$⟹\alpha \delta =\beta \gamma =\frac{r}{p}\left[\text{Sum of the roots is}p\right]$

$\alpha \beta \gamma \delta =\left(\frac{r}{p}\right)\left(\frac{r}{p}\right)=s\left[\text{Product of the roots is}s\right]$

$⟹r^2=p^{2}s$