What should be the relationship between $p$ and $q$, so that $x^4+px^2+q=0$ equations has four solutions which are successive members of arithmetic equation.
The answer is root from $q / p = -3/10$, but I have no clue what is going on here and would like to know. All i could deduce before seeing the answer is that $p<0$ and $q>0$. I would appreciate if someone who can solve this replies and helps. Thanks
Best Answer
Ignoring the case $p=q=0$, the solutions to this equation will be of the form $x=\pm \alpha, \pm \beta \ (\alpha\gt\beta \gt 0)$ and the arithmetic sequence (in order) would be $$-\alpha, -\beta,\ \beta, \alpha$$ By definition of an arithmetic progression, $$\alpha -\beta = 2\beta \implies \alpha =3\beta$$ Applying Vieta’s formulas on $x^4+px^2+q=0$, $$\alpha^2+\beta^2 = -p \\ \alpha^2 \beta^2 = q $$$$\implies10\beta^2 =-p \\ 3\beta^2 =\sqrt q $$ or $$\frac{\sqrt q}{p} =\frac{-3}{10}$$