Bounding roots of a parabola

Question

Given the quadratic equation $$ax^2+bx+c=0,$$ I am trying to bound its roots. The solutions are $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ but using the triangle inequality $$|x_{1,2}|\leq\frac{|b|+\sqrt{b^2-4ac}}{2|a|}=\frac{|b|+\sqrt{(b-2\sqrt{ac})(b+2\sqrt{ac})}}{2|a|}$$ and by AM-GM inequality we have $$\frac{|b|+\frac{(|b|-2\sqrt{ac})+(|b|+2\sqrt{ac})}2}{2|a|}=\frac{|b|+|b|}{2|a|}=\left|\frac{b}{a}\right|.$$ Thus $$|x_{1,2}|\leq \left|\frac{b}{a}\right|,$$ However, if I try this with $3x^2-x-19=0$, I have that one root is $2.68$, while $b/a=1/3$. Why is this wrong?

Answer 1

This answer is assuming that you're operating under the restriction that the coefficients and the roots are real. What you're doing is basically saying that:

$$\sqrt{b^2-4ac} \leq |b|$$

which is obviously false if $ac < 0$.

The error in the AM-GM application (which can only be applied for non-negative numbers) stems from assuming $\sqrt{ac}$ exists, which it doesn't when $ac < 0$, as is the case of your equation.

The bound is good however, for second degree equations with $\frac{b^2}{4} \geq ac \geq 0$.