Problem
Prove that all roots of $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real. Here $a,b \in \mathbb R$, and $a \neq 0$.
Source
This is one of the previous year problem of Regional Math Olympiad (India). I had a hard time solving it, so thought I'd better ask here.
Observations
- Some real roots are possible: when $a<0$, the equation has two of them.
- If one more coefficient was allowed to be arbitrary: $a x^4 + b x^3 + cx^2 + x + 1 = 0$, then the roots could be all real, since every quartic can be brought into such form by scaling
Best Answer
Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic $ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots iff $f$ does. But then the same is true of $f(x-\frac14) = x^4 + \frac58 x^2 + Bx + A$ for some $B$ and $A$ (we don't need the formula). If this is $(x-a)(x-b)(x-c)(x-d)$ for some $a,b,c,d$ then $a^2+b^2+c^2+d^2 = -2\frac58 < 0$, so $a,b,c,d$ cannot all be real, QED.