# How does $\left|\sqrt{2+3x_1} - \sqrt{2+3x_2}\right|$ become $\left| \frac{3(x_1-x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}} \right|$?

Could someone explain to me this transformation? It is used frequently in my uni course, and I do not understand what's happening:

$$\left|\sqrt{2+3x_1} - \sqrt{2+3x_2}\right| = \left| \frac{3(x_1-x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}} \right|\qquad (x_1,x_2 \in [0,3))$$

Is there some obvious rule I don't see? Thanks in advance. (The transformation is used to prove Lipschitz continuity.)

• If $A_i = \sqrt{2+3x_i}$, then what is $(A_1-A_2)(A_1+A_2)$? – Don Thousand Jun 3 '19 at 2:58

Let $$A_i = \sqrt{2+3x_i}$$ for $$i=1,2$$ as suggested by Don Thousand in the comments. Then your original expression is $$|A_1 - A_2|$$.

Notice, then:

$$A_1 - A_2 = (A_1-A_2) \cdot \underbrace{\frac{A_1 + A_2}{A_1 + A_2}}_{=1} = \frac{A_1^2 - A_2^2}{A_1+A_2}$$

$$A_i^2 = 2+3x_i$$ for both $$i$$ indices and thus the difference in the numerator is $$2+3x_1 - 2 - 3x_2$$, or, more simply, $$3(x_1 - x_2)$$.

Then

$$A_1 - A_2 = \cdots = \frac{A_1^2 - A_2^2}{A_1+A_2} = \frac{3(x_1 - x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}}$$

Throw in the absolute values and you're done!