How to do a quick estimation if $x_2 \ll x_1$ holds for the roots of a quadratic equation - to apply quick and easy root-finding formula? Wikipedia provides an interesting method of (approximately) solving a quadratic equation:

Vieta's formulas provide a useful method for finding the roots of a
quadratic in the case where one root is much smaller than the other.
If $|x_2|\ll|x_1|$, then $x_1+x_2\approx x_1$ and we have the estimate $$x_1\approx-\frac ba$$

But how can I quickly estimate that one root is much larger than the other?
 A: One root is much smaller than the other when $|ac| \ll b^2$ because then the square root in the quadratic formula is very close to $b$.  The approximation given comes from replacing the square root by $b$ and taking the minus sign so the two terms add.  This is also the time that the calculation of the other root suffers from roundoff error in a computer because $-b+\sqrt{b^2-4ac}$ is the difference of two similar numbers.
A: Another derivation of the first answer...
Let one of the roots be $x_1$ and let the other root be $x_2=kx_1$.
Then sum of roots is $x_1+kx_1=(k+1)x_1$ and product of roots is $x_1 \times kx_1=kx_1^2$
For $ax^2+bx+c=0$ we have $-\frac ba=(k+1)x_1$ and $\frac ca=kx_1^2$
Square the first so that $\frac {b^2}{a^2}=(k+1)^2x_1^2 \Rightarrow \frac {b^2}{ac}=\frac{(k+1)^2}{k}$
For $|k| \gg 1, \frac {b^2}{ac} \approx k+2 \Rightarrow |\frac {b^2}{ac}| \gg 1$
For $|k| \ll 1, \frac {b^2}{ac} \approx \frac 1k \Rightarrow \frac {a^2}{bc} \approx k \Rightarrow |\frac {ac}{b^2}| \ll 1 \Rightarrow |\frac {b^2}{ac}| \gg 1$
In either case, a large value of $|\frac {b^2}{ac}|$ means that one root is much larger than the other.
