$x^4-4x^3+ax^2+bx+1=0$ What should $a$ and $b$ be, so that the given equation has four real roots?

  1. $(4, -6)$
  2. $(6, -4)$

Not getting any start. Tried Descartes sign rule conditions to real roots but couldn't figure out the exact answer.

  • $\begingroup$ Do you mean ONLY real roots ? or at least one real root? $\endgroup$ – SchrodingersCat Nov 5 '15 at 14:15

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 $x = 1$.

  • $\begingroup$ so whats the answer for it $\endgroup$ – Archis Welankar Nov 5 '15 at 14:15
  • $\begingroup$ @ArchisWelankar Clarify the question! $\endgroup$ – Jean-Claude Arbaut Nov 5 '15 at 14:16
  • $\begingroup$ all 4 an edit has been made @jean $\endgroup$ – Archis Welankar Nov 5 '15 at 14:20
  • $\begingroup$ @ArchisWelankar This answer as written still addresses the edited version of the equation, no? $\endgroup$ – Travis Willse Nov 5 '15 at 14:22
  • 1
    $\begingroup$ I'm not sure what the confusion is, exactly: The answer says that there are precisely two real solutions for $(a, b) = (4, -6)$, and one can show this using the method I mentioned (showing that the derivative of the l.h.s. only has a single root). $\endgroup$ – Travis Willse Nov 5 '15 at 14:33

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