What math technique is used to get $k\left(\sqrt{r_o^2+x^2}-r_o\right) \approx kr_o \left(1+\frac12 \frac{x^2}{r_o^2}-1\right)$?

Question

How is this equation in left-hand side, approximately equal to the right hand side? What math technique is used? $$k\left(\sqrt{r_o^2+x^2}-r_o\right) \approx kr_o \left(1+\frac12 \frac{x^2}{r_o^2}-1\right)$$

If you need more context, here it is: this is from electrical engineering; electromagnetism; antenna arrays;

Question:

Answer:

In part b, I need to find a maximum distance limit where phase error of antenna beam would be valid. I understand everything until the eqn. they got from approximating.

I have all the information, there is no more info to add.

Answer 1

This is the Taylor series of $\sqrt{1+x}$ which is equal to $$ \sqrt{1+x}=1+\frac12x-\frac18x^2+... $$ For small values of $x$ the terms $x^2$ etc. can be discarded.

Answer 2

\begin{align} k\left(\sqrt{r_o^2+x^2}-r_o\right) &= k\left(\sqrt{r_o^2\left(1+\frac{x^2}{r_o^2}\right)}-r_o\right) \\ &= k\left(r_o \sqrt{1+\frac{x^2}{r_o^2}}-r_o\right) \\ &= kr_o \left(\sqrt{1+ \frac{x^2}{r_o^2}}-1\right) \\ &\approx kr_o \left(1+\frac12 \frac{x^2}{r_o^2}-1\right). \end{align}

The approximation from the last step comes from this Taylor series: $$\sqrt{1+x} = 1 +\frac12 x -\frac18 x^2 +\frac1{16} x^3 -\frac5{128} x^4 +\frac7{256} x^5 +\ldots$$ For small $x$, values like $x^2$, $x^3$, et cetera, can be neglected.