The problem: Find the values of $m$ for which all $4$ roots of the polynomial $(2m-1).x^4-2mx^2+1=0$ make an arithmetic progression. Then Find the roots for the found values of $m$.
What I've done: All $4$ to exist $2m-1$ must NOT $=0$ so $m =0,5$ not an answer. Also the $D>0$ $(m-1)^2>0$ so $1$ is not an answer. The roots of the polynomial are $1,-1,-\sqrt{-1/2m-1},\sqrt{-1/2m-1}$
Then for them to be an arithmetic progression each must = (the sum of other 2)/2 ? So I must solve some equations like : $1=(-1+\sqrt{-1/2m-1})/2$?
Hint: the equation is a biquadratic, so if $x_1$ is a root then $-x_1$ is also a root. Then the $4$ roots must be of the form $-3r,-r,r,3r$ for some $r \gt 0\,$.
By Vieta's formulas the product of the $4$ roots is $9r^4=\frac{1}{2m-1}\,$, so $r = \frac{1}{\sqrt{3}\,\sqrt[4]{2m-1}}$. Substituting this value in the equation gives a quadratic in $m\,$, with two real roots which both satisfy the condition.