Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would you simplify the following radical.

$$\sqrt\frac1B$$

I am kind of confused do I multiply the numerator and denominator by B.

share|improve this question

2 Answers 2

$\sqrt{\frac{1}{B}} = \sqrt{\frac{1}{B}\frac{B}{B}} = \sqrt{\frac{B}{B^2}} = \frac{\sqrt{B}}{\sqrt{B^2}} = \frac{\sqrt{B}}{B}$

of course assuming that $B \neq 0$.

Although I am not sure which is more simple or appealing.

share|improve this answer
    
You need $B>0$ because it's inside the root, merely $B\neq 0$ is not enough. –  Asaf Karagila Jun 13 '12 at 20:43
    
@William: Agree with your comment on simplification. The "unsimplified" version can be approximated on the calculator by entering $B$, pressing reciprocal, pressing square root. The "simplified" version takes more keystrokes. A form that is simple for one purpose may not be so simple for another. –  André Nicolas Jun 13 '12 at 20:51

$$\sqrt\frac1B=\frac1{\sqrt B}=\frac1{\sqrt B}\cdot\frac{\sqrt B}{\sqrt B}=\frac{\sqrt B}{(\sqrt B)^2}=\frac{\sqrt B}B$$

Of course that we have to have $B>0$ for this to be meaningful in the context of real numbers.

share|improve this answer
1  
You mention both of what I would consider "simpler" forms: $\dfrac{1}{\sqrt{B}}$ and $\dfrac{\sqrt{B}}{B}$; however, I see nothing complicated about $\sqrt{\dfrac1B}$. (+1) –  robjohn Jun 13 '12 at 21:03
    
@robjohn: I completely agree. However at a precalc level I could see why $\sqrt\frac1B$ may seem a bit complicated. –  Asaf Karagila Jun 13 '12 at 21:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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