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{a^{2}+b} &= a\sqrt{1+\frac{b}{a^{2}}} \\
  & \approx a\left( 1+\frac{b}{2a^2} \right) \\
  &= a+\frac{b}{2a}
\end{align*}

Now
\begin{align*}
  \sqrt{10^{2}-(6.9)^{2} \times 10^{-4}}
  & \approx 10-\frac{(6.9)^{2} \times 10^{-4}}{20} \\
  &= 10-\frac{(6.9)^{2} \times 10^{-5}}{2} \\
  & \approx  10-0.000238 \\
  & = 9.999762
\end{align*}
