Show that $x(1- (\frac{R_1-R_2}{x})^{2})^{0.5}$ = x – $\frac{(r1 – r2)^2}{2x}$

Question

Show that $x(1- (\frac{R_1-R_2}{x})^{2})^{0.5}$ = x [1 – {0.5 $\frac{(r1 – r2)}{x}$ + … ] = x – $\frac{(r1 – r2)^2}{2x}$ by using binomial theorem it was mentioned that binomial theorem is $(x+y)^2$ then $x^2$+$2xy$+$y^2$ but if i apply this to $(r_1-r_2)^2$ then raised again to a half wait how?

Answer 1

$$\left\{1-\left(\dfrac{R_2-R_1}x\right)^2\right\}^{1/2}$$

$$=1-\dfrac12\cdot\left(\dfrac{R_2-R_1}x\right)^2+\text{ terms of higher powers of }\left(\dfrac{R_2-R_1}x\right)^2$$