Approximation of square root I have a square root in a problem which needs to be approximated. I'm not entirely sure how to do this algebraically.
$$ \sqrt{10^2-(6.9\times 10^{-2})^2}$$
The answer the problem is proposing as the approximation is $$ 10[1-\frac 1 2(6.9)^2\times10^{-6}]$$
This hasn't exactly been the most reputable textbook, however, so it could be wrong.

My attempt:
It should be able to approximated by the square-root of 100, because the other number is so small.
$$ 10 -(6.9)^2\times 10^{-4} $$
I can see they factored a 10 out, so I go ahead and do that.
$$ 10[1- \frac 1 {10}(6.9)^2\times 10^{-4}] $$
$$ 10[1-(6.9)^2\times 10^{-5}]  $$
I'm not sure where exactly I'm slipping up, or what I'm comprehending incorrectly, any help would be greatly appreciated.

$$\begin{align} \\ & \sqrt{10^2-(6.9\times 10^{-2})^2}
\\ & = \sqrt{10^2\left[1 -\frac{1}{10^{2}}\left\{(6.9)^2\times 10^{-4}\right\}\right]}
\\ & = \left[10^2\{1 -(6.9)^2\times 10^{-6}\}\right]^{\frac{1}{2}}
\\ & \approx 10\left[1-\frac 1 2(6.9)^2\times10^{-6}\right] \,\,\,\,\,\,\,\,\,\,\,\,\, \text{using binomial approximation}  \end{align}$$ The  binomial approximation is stated in this link.
