# Checking whether there are real solutions to the polynomial equation $x^4-4x^3+ax^2+bx+1=0$ for certain $a, b$

$x^4-4x^3+ax^2+bx+1=0$ 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.

• Do you mean ONLY real roots ? or at least one real root? – 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 $x = 1$.
• 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). – Travis Willse Nov 5 '15 at 14:33