1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Try adding the terms under square root taking lcm $(=4x^2)$ of the denominator $\endgroup$ Jun 20, 2013 at 18:34
  • 2
    $\begingroup$ The most important fact is that $\sqrt{x^2}=|x|$ NOT $x$. $\endgroup$
    – Sigur
    Jun 20, 2013 at 18:36

2 Answers 2

1
$\begingroup$

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 $x>0$.

$\endgroup$
1
  • 1
    $\begingroup$ Just to make it really explicit: The properties of square roots that you're using are: $$\sqrt{a\cdot b} = \sqrt{a}\cdot\sqrt{b}$$ $$\sqrt{a^2} = a\quad\text{(iff $a \ge 0$)}$$ $\endgroup$
    – apnorton
    Jun 20, 2013 at 18:41
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ It indeed does in the problem I encountered it, forgot to mention. However, still don't understand this $\endgroup$ Jun 20, 2013 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.