<s>[INST] 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 $c$?
 [/INST] 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 $x_2$.
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