Solutions of biquadratic equation being successive members of arithmetic progression

Question

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

Answer 1

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 $$\frac{\sqrt q}{p} =\frac{-3}{10}$$

Answer 2

Let's assume that you want 4 distinct solutions. You notice that the equation contains only terms in $x^0, x^2, x^4$. Therefore if $x_0$ is a solution then $-x_0$ is a solution as well. You have four solutions: $-x_2<-x_1<x_1<x_2$. You also know that the distance between solutions is constant and equal to $2x_1$. That means $x_2=3x_1$. Now write $p$ and $q$ in terms of $x_1$. Eliminate $x_1$ between the two equations, to get the relationship between $p$ and $q$.

Answer 3

You want the roots of your polynomial to be $a$, $a+r$, $a+2r$, $a+3r$ for some $r$.
But $x^4 + p x^2 + q$ is an even function of $x$, thus the roots must be symmetric about $0$, which means $a = -(a+3r)$ and $a+r = -(a+2r)$. This is equivalent to $r = - 2a/3$. Then the roots are $a, a/3, -a/3, -a$, and the polynomial is $$(x-a)(x-a/3)(x+a/3)(x+a) = (x^2 - a^2)(x^2 - a^2/9) = x^4 - \frac{10}{9} a^2 + \frac{a^4}{9}$$ which is $x^4 + p x + q$ with $p = -10 a^2/9$ and $q = a^4/9$. The relation is $q = \dfrac{9}{100} p^2$, with $p \le 0$ if you want the roots to be real.

Answer 4

A biquadratic polynomial can be factored as the product of two quadratic polynomials. Because the polynomial function has even symmetry, its zeroes are arranged symmetrically about the $ \ y-$axis as $ \ -s \ , \ -r \ , \ r \ , s \ \ . $ If we call $ \ S \ = \ r + s \ $ and $ \ P \ = \ r·s \ \ , $ the Viete relations tell us that the factorization is $$ x^4 \ + \ p·x^2 \ + \ q \ \ = \ \ (x^2 \ + \ S·x \ + \ P)·(x^2 \ - \ S·x \ + \ P) $$ $$ = \ \ x^4 \ + \ (2P - S^2)·x^2 \ + \ P^2 \ \ . $$

Since the zeroes are to form an arithmetic progression, we must have $$ \ -r \ = \ -s + d \ \ , \ \ r \ = \ -s + 2d \ \ , \ \ s \ = \ -s + 3d \ \ \Rightarrow \ \ s \ = \ \frac32·d \ \ , \ \ r \ = \ \frac12·d $$ $$ \Rightarrow \ \ S \ = \ 2d \ \ , \ \ P \ = \ \frac34·d^2 \ \ , $$ which makes the coefficients of the biquadratic polynomial $ \ p \ = \ 2·\frac34·d^2 \ - \ (2d)^2 \ = \ -\frac{5}{2}·d^2 \ $ and $ \ q \ = \ (\frac34·d^2)^2 \ = \ \frac{9}{16}·d^4 \ \ . $ Hence, $$ \frac{\sqrt{q}}{p} \ \ = \ \ \frac{P}{-\frac{5}{2}·d^2} \ \ = \ \ \frac{\frac{3}{4}·d^2}{-\frac{5}{2}·d^2} \ \ = \ \ \frac34 \ · \ \left(-\frac25 \right) \ \ = \ \ -\frac{3}{10} \ \ . $$