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.

Use the Taylor series for $(1+x)^{1/2}$.
First we rewrite $(A+B)^{\alpha}$ to be in a form similar to this : 
$(A+B)^{\alpha} = A^{\alpha}(1 + \frac{B}{A})^{\alpha}$. 
With $A \gg B$ the ratio $\frac{B}{A}$ is "small" and we do a Taylor series for $(1+x)^{\alpha}$ 
$$(1+x)^{\alpha} = 1 + \frac{\alpha}{1!}x + \frac{\alpha(\alpha-1)}{2!}x^{2} + \ldots$$
and now replace each $x$ with $\frac{B}{A}$ to get
$$(A+B)^{\alpha} = A^{\alpha}\left( 1 + \frac{\alpha}{1!} \frac{B}{A} + \frac{\alpha(\alpha-1)}{2!}\left( \frac{B}{A} \right)^{2} + \ldots \right)$$
Putting in $\alpha = \frac{1}{2}$ gives
$$(A+B)^{\frac{1}{2}} = A^{\frac{1}{2}}\left( 1 + \frac{1}{2} \frac{B}{A} - \frac{1}{8}\left( \frac{B}{A} \right)^{2} + \ldots \right)$$
In your case $A = 10^{2}$ and $B = -(0.69)^{2}$ so $A^{\frac{1}{2}} = 10$ and $\frac{B}{A} = -(0.069)^{2} = -(6.9 \times 10^{-3})^{2} = -0.00004761...$
Therefore
$$\begin{align}
(A+B)^{\frac{1}{2}} &= \sqrt{10^{2}}\Big( 1 + \frac{1}{2}\frac{-(0.69)^{2}}{10^{2}} + \ldots \Big) = 10\left( 1 - \frac{0.00004761}{2} + \ldots \right) 
\\&\approx 10*(1-0.000023805) = 9.99976195
\end{align}$$
