$x^4-4x^3+ax^2+bx+1=0$ What should $a$ and $b$ be, so that the given equation has four real roots?
- $(4, -6)$
- $(6, -4)$
Not getting any start. Tried Descartes sign rule conditions to real roots but couldn't figure out the exact answer.
When $(a, b) = (4, -6)$, evaluating the l.h.s. at $x = 1$ gives $-4$, but the leading coefficient is positive, so there are at least two distinct, real solutions. (In fact, we can show that the derivative of the l.h.s. has a single root, so there are precisely two real solutions.)
When $(a, b) = (6, -4)$, the l.h.s. factors as $(x - 1)^4$, so the equation has a (unique) solution at $x = 1$.