Question

If the roots of the equation $x^4+p{x}^3+q{x}^2+rx+s=0$ are in arithmetical progression, show that $p^3-4pq+8r=0$; and if they are in geometrical progression, show that $p^{2}s=r^2$.

Answer 1

Given, $x^4+p{x}^3+q{x}^4+rx+s=0$

Case 1: Roots are in AP. Let the roots be $\alpha -3\delta ,\alpha -\delta ,\alpha +\delta ,\alpha +3\delta$

Sum of the roots, $S_1=4\alpha =-p⟹\alpha =\frac{-p}{4}\dots \dots \dots \dots .\left(1\right)$

Sum of the roots taken 2 at a time, $S_2=6{\alpha }^2-10{\delta }^2=q\dots \dots \dots \dots .\left(2\right)$

Sum of the roots taken 3 at a time, $S_3=4{\alpha }^3-20\alpha {\delta }^2=-r\dots \dots \dots \dots .\left(3\right)$

Multiplying (2) with $2\alpha$ and Subtracting from (3), we get

$8{\alpha }^3=2q\alpha +r⟹\frac{-p^3}{8}=\frac{-2pq}{4}+r⟹p^3-4pq+8r=0$

Hence Proved

Case 2: Roots are in GP. Let the roots be $\frac{\alpha }{{\delta }^2},\frac{\alpha }{\delta },\alpha \delta ,\alpha {\delta }^2$

Sum of the roots, $S_1=\alpha (\frac{1}{{\delta }^2}+\frac{1}{\delta }+\delta +{\delta }^2)=-p\dots \dots \dots \dots .\left(1\right)$

Sum of the roots taken 3 at a time, $S_3={\alpha }^3(\frac{1}{{\delta }^2}+\frac{1}{\delta }+\delta +{\delta }^2)=-r\dots \dots \dots \dots .\left(2\right)$

Product of the roots, $S_4={\alpha }^4=s\dots \dots \dots \dots .\left(3\right)$

From (1) and (2), we have

${\alpha }^2\cdot -p=-r$

Taking square on both side and substituting value of ${\alpha }^4$ from (3), we have

$p^{2}s=r^2$

Hence Proved