Quadratic equation estimating changes in b and c If $x^2 + bx + c=0$ is an equation the roots are
$$
\begin{aligned}
x&=\frac{-b + \sqrt{b^2 - 4c}}{2}\\
x&=\frac{-b - \sqrt{b^2 - 4c}}{2}
\end {aligned}$$
What is the effect of small changes in $b$ or $c$?
For example if I have $\sqrt{a}$ then to estimate a small change in this equation I would write
$$\sqrt{a + \Delta a}$$
and set $v = \sqrt{a}$ and 
$$v + \Delta v =\sqrt{a + \Delta a}$$
Therefore $a + \Delta a = (v + \Delta v)^2$ which is roughly
$v^2 + 2v \times \Delta v$ (ignoring $\Delta v^2$ as it's too small) and thus
$$\Delta a = 2\sqrt{a} \times \Delta v$$
so with some rearranging $\Delta v$ is roughly $\frac{1}{2\sqrt{a}} \times\Delta a$
(to the first order) and this lets me estimate square roots.
How can I use a method like this to estimate small changes in $b$ and $c$?
 A: In general you should consider the function 
$$f(b,c)=\frac{-b + \sqrt{b^2 - 4c}}{2}$$
and
$$f(b,c)=\frac{-b - \sqrt{b^2 - 4c}}{2}$$
and calculate partial derivatives $f_b$, $f_c$ and gradient $\Delta f=(f_b,f_c)$.
For every small change $v=(\delta b, \delta c)$ an extimation of the rate of change of the solution is
$$\Delta f \cdot v$$
The nice thing is that by the gradient you are able to determine for which “directions” $v$ the rate of change is maximum ($\Delta f$ and $v$ parallel) and minimum ($\Delta f \cdot v$=0).
A: Let us consider $$x_1=\frac{-b + \sqrt{b^2 - 4c}}{2} \qquad \text{and} \qquad x_2=\frac{-b - \sqrt{b^2 - 4c}}{2}$$ and use Taylor expansions around, say, $b_0$ or $c=c_0$.
For a change in $b$, what we would obtain is 
$$x_1=\frac{1}{2} \left(\sqrt{b_0^2-4 c}-b_0\right)+(b-b_0) \left(\frac{b_0}{2 \sqrt{b_0^2-4
   c}}-\frac{1}{2}\right)+O\left((b-b_0)^2\right)$$
$$x_2=-\frac{1}{2}\left( \sqrt{b_0^2-4 c}+{b_0}\right)+(b-b_0) \left(-\frac{b_0}{2
   \sqrt{b_0^2-4 c}}-\frac{1}{2}\right)+O\left((b-b_0)^2\right)$$ that is to say
$$\Delta x_{1,2}=\left(\frac{b_0}{2 \sqrt{b_0^2-4
   c}}-\frac{1}{2}\right)\Delta b$$
For a change in $c$, what we would obtain is 
$$x_1=\frac{1}{2} \left(\sqrt{b^2-4 c_0}-b\right)-\frac{c-c_0}{\sqrt{b^2-4
   c_0}}+O\left((c-c_0)^2\right)$$
$$x_2=-\frac{1}{2} \left(\sqrt{b^2-4 c_0}+{b}\right)+\frac{c-c_0}{\sqrt{b^2-4
   c_0}}+O\left((c-c_0)^2\right)$$ that is to say
$$\Delta x_{1,2}=\pm {\sqrt{b^2-4
   c_0}}\,\Delta c$$
Edit
I did focus too much on the quadratic equation. So, let us do it for a general equation $$F(x,a_1,a_2,\cdots,a_n)=0$$ (considered as an implicit function) and let compute the partial derivatives. We have 
$$\frac{\partial x}{\partial a_i}=-\frac{\frac{\partial F(x,a_1,a_2,\cdots,a_n)}{\partial a_i}} {\frac{\partial F(x,a_1,a_2,\cdots,a_n)}{\partial x}}$$ which makes $$\Delta x=\frac{\partial x}{\partial a_i}\Delta a_i$$ which can be applies to any root (analytical or not) and any parameter.
Applied to $F(x,b,c)=x^2+bx+c=0$, this would lead to $$\Delta x=-\frac{x}{b+2 x} \Delta b \qquad \text{and}\qquad \Delta x=-\frac{1}{b+2 x}\Delta c$$ Now, replace $x$ by $x_1$ and $x_2$.
A: This is know as Taylor expansion.
When $x\ll 1$ is small then $(1+x)^\alpha=1+\alpha x+(\alpha)(\alpha-1)\dfrac{x^2}{2!}+\alpha(\alpha-1)(\alpha-2)\dfrac{x^3}{3!}+\cdots$
For $\alpha=\frac 12$ then $\sqrt{1+x}=1+\dfrac x2-\dfrac{x^2}8+\cdots$
This is why you find $\sqrt{a+\Delta a}=\sqrt{a}\sqrt{1+\frac{\Delta a}a}\approx\sqrt{a}(1+\dfrac{\Delta a}{2a}))=\sqrt{a}+\dfrac{\Delta a}{2\sqrt{a}}$ at first order, like you discovered intuitively.
Here is a small compendium of common Taylor expansion in a neighbourhood of zero: 
http://www.h-k.fr/publications/data/adc.ps__annexes.maths.pdf
You can then find the variations of the roots for small changes on $b$ and $c$, and that lead to some complicated formulas.

You may want also to consider that in $x^2-sx+p=0$ we have $s=x_1+x_2$ and $p=x_1x_2$.
Since the variations on $s=-\dfrac b{a^2}$ and $p=\dfrac c{a^2}$ are easy to calculate, you may approximate the variations on the roots when they are of the same order by condering $\Delta x_1\approx\Delta x_2\approx\Delta x$.
$s+\Delta s\approx x_1+x_2+2\Delta x$ and having $\Delta x\approx\frac 12\Delta s\approx-\dfrac {\Delta b}{2a^2}$
And similarly for the product $\Delta x\approx \dfrac{\Delta p}s\approx -\dfrac{\Delta c}b$

Of course in the case the roots are not of the same order (for instance $0.1$ and $10$), this approximation fails and you may want to consider individual approximation instead.
