Bump!
Question: Find the minimum possible value of such that the quartic polynomial has at least 1 real root.
Attempt: Since the equation is palindromic, the standard technique is:
Since the real roots must be 1 or -1 correct? But when I solve simultaneous equations using 1 and -1 as roots I get which is too large. The glitch I ran into was excluding the case that at the minimal value of there might be two real roots of multiplicity 2. That made things a little ugly.
I also tried this approach: Suppose r is a simple multiplicity 1 root and a^2+b^2 is at it's minimal value. The graph of the polynomial near the root the just looks like a line crossing the x-axis, right? That means if I move the graph just a little by varying a and b, there will still be a real root because the graph will still cross the x-axis, just in a slightly different spot. That in turn means I can make a or b just a little smaller so I decrease and still have a root. So that would contradict being the smallest value where you still have a real root.
That means at the minimal value, all of the real roots must have multiplicity 2 or greater but I think I am stuck here...