Problem - 3207
Show that if $a, b, c$ are the lengths of the sides of a triangle, then the equation $$b^2x^2 +(b^2+c^2-a^2)x+c^2=0$$ does not have real roots.
By law of cosines, we have $b^2+c^2 -a^2 = 2bc\cos A$. Therefore, the given equation is equivalent to
$$b^2 x^2 + 2bc\cos A x + c^2=0$$
Its determinant equals
$$(2bc\cos A)^2 - 4\cdot b^2\cdot c^2=4b^2c^2(\cos^2 A - 1) \le 0$$
However the equality cannot hold because $0 < A < 180^\circ$ which implies $\cos^2 A < 1$.