# [SOLVED]Root of Quartic Polynomial

#### anemone

##### MHB POTW Director
Staff member
Let $a$ and $b$ be real numbers such that $a\ne 0$. Prove that not all the roots of $ax^4+bx^3+x^2+x+1=0$ can be real.

#### Opalg

##### MHB Oldtimer
Staff member
Between each pair of real roots of a polynomial there must be a root of the derivative.

Let $x_1,x_2,x_3,x_4$ be the roots of $ax^4+bx^3+x^2+x+1$. Replacing $x$ by $\frac1x$, it follows that $\frac1{x_1},\frac1{x_2},\frac1{x_3},\frac1{x_4}$ are the roots of $p(x) = x^4 + x^3 + x^2 + bx + a$. The second derivative of $p(x)$ is $p''(x) = 12x^2 + 6x + 2$, which has no real roots. So $p'(x)$ can have only one real root, and $p(x)$ has at most two real roots. Therefore at most two of $x_1,x_2,x_3,x_4$ are real.