# Algebraic manipulation with square roots

I have always had problems with the algebraic manipulation of square roots. For example, recently I encountered this in a problem I was working on:

$$\sqrt{\left(\dfrac{x-1}{2x}\right)^2 - \dfrac{y}{x}} = \dfrac{1}{2x} \sqrt{(x-1)^2 -4xy}$$

I still don't grasp why this is correct and in general, I have trouble knowing when you can factor out something when dealing with square roots. Can someone enlighten me.

• Try adding the terms under square root taking lcm $(=4x^2)$ of the denominator – lab bhattacharjee Jun 20 '13 at 18:34
• The most important fact is that $\sqrt{x^2}=|x|$ NOT $x$. – Sigur Jun 20 '13 at 18:36

Observe that: \begin{align*} \sqrt{\left(\dfrac{x-1}{2x}\right)^2 - \dfrac{y}{x}} &= \sqrt{\dfrac{(x-1)^2}{(2x)^2} - \dfrac{y}{x} \cdot \dfrac{4x}{4x}} \\ &= \sqrt{\dfrac{(x-1)^2}{(2x)^2} - \dfrac{4xy}{4x^2}} \\ &= \sqrt{\dfrac{(x-1)^2}{(2x)^2} - \dfrac{4xy}{(2x)^2}} \\ &= \sqrt{\dfrac{(x-1)^2-4xy}{(2x)^2}} \\ &= \sqrt{\dfrac{1}{(2x)^2} \cdot \left( (x-1)^2-4xy \right)} \\ &= \sqrt{\left(\dfrac{1}{2x}\right)^2 \left((x-1)^2 -4xy\right)} \\ &= \sqrt{\left(\dfrac{1}{2x}\right)^2}\cdot \sqrt{(x-1)^2 -4xy} \\ &= \dfrac{1}{2x} \sqrt{(x-1)^2 -4xy} \end{align*} assuming that $x>0$.
• Just to make it really explicit: The properties of square roots that you're using are: $$\sqrt{a\cdot b} = \sqrt{a}\cdot\sqrt{b}$$ $$\sqrt{a^2} = a\quad\text{(iff a \ge 0)}$$ – apnorton Jun 20 '13 at 18:41
That's not always true. In fact, $$\sqrt{\left(\dfrac{x-1}{2x}\right)^2 - \dfrac{y}{x}} = \dfrac{1}{2|x|} \sqrt{(x-1)^2 -4xy}$$
If domain of $x$ contains only positive values then you equality is valid
EDIT: $$\sqrt{\left(\dfrac{x-1}{2x}\right)^2 - \dfrac{y}{x}}=\sqrt{\dfrac{(x-1)^2}{4x^2} - \dfrac{y}{x}}=\sqrt{\dfrac{(x-1)^2-4xy}{4x^2}}=\dfrac{1}{2|x|} \sqrt{(x-1)^2 -4xy}$$