Inspired by an IMO problem!

The polynomial in the left side of the equation can be factored as follows: $x^4+ax^3+bx^2+ax+1=(x^2+px+r)(x^2+qx+1/r)\:\:\:\:\:\:\:(*)$ where $r=\alpha*\beta$ and $p=-(\alpha+\beta)$ and $q$ is a real number.

Clearly, $r\not=1$ . From the equality (*) by expanding the product and setting corresponding coefficients equal, we get the following expressions for $a$ and $b$ in terms of $p$ , $q$ and $r$ $a=p+q= \frac{p}{r}+qr$ $b=r+\frac{1}{r}+pq.$

Since $r\not=1$ , the equality $p+q=\frac{p}{r}+qr$ implies that $p=rq$ . Now , as the zeros of the quadratic factor $x^2+px+r$ are real, then $p^{2}\ge 4r.$ . Dividing both sides of this inequality by $r^2$ we obtain that $q^2\ge\frac{4}{r}$ and therefore the zeros of the quadratic factor $(x^2+qx+1/r)$ are also real. Thus we can conclude that all the solutions of the given quartic equation are real.