Algebra with surds

[dariel]

I'm following a worked example for an electronics problem, but can't see how my lecturer got from one step to the next, most of the people in the class have good algebra so he doesn't explain. Usually I can get it after the class but this one has me stuck. I'm not sure of the method to use on the surd.

If anyone can explain I'd really appreciate it.

[MATH] \frac{-RC \pm \sqrt{R^2C^2-4LC}}{2LC}=\frac{-R}{2L} \pm \frac{R}{2L} \sqrt {1-\frac{4L}{R^2C}}[/MATH]
The part that gets me is getting the [MATH]\frac{R}{2L}[/MATH] out in front of the surd I think. I've managed to do the other manipulations fine.

 

[pka]

I'm following a worked example for an electronics problem, but can't see how my lecturer got from one step to the next, most of the people in the class have good algebra so he doesn't explain. Usually I can get it after the class but this one has me stuck. I'm not sure of the method to use on the surd. If anyone can explain I'd really appreciate it.
[MATH] \frac{-RC \pm \sqrt{R^2C^2-4LC}}{2LC}=\frac{-R}{2L} \pm \frac{R}{2L} \sqrt {1-\frac{4L}{R^2C}}[/MATH]The part that gets me is getting the [MATH]\frac{R}{2L}[/MATH] out in front of the surd I think. I've managed to do the other manipulations fine.

Please tell us which of these is true: \(\displaystyle R>0,~L>0,\text{ or }C>0\)

 

[dariel]

Ah, apologies, LHS = 0, and RHS is the re-arrangement of the left, all are > 0

 

[pka]

Please tell us which of these is true: \(\displaystyle R>0,~L>0,\text{ or }C>0\)

Thank you, otherwise it ain't true.

\(\displaystyle \sqrt{R^2C^2 - 4LC} = \sqrt{{R^2}{C^2} - \frac{{4L{C^2}{R^2}}}{{C{R^2}}}} \\
= RC\sqrt {1 - \frac{{4L}}{{C{R^2}}}}\)

 

[JeffM]

i am going to do this step by step. (It is just a longer way to do what pka did.)

[MATH]\dfrac{-\ rc \pm \sqrt{r^2c^2 - 4kc}}{2kc} =[/MATH]
[MATH]\dfrac{-\ rc \pm \sqrt{r^2c^2 - \dfrac{4kc^2}{c}}}{2kc} =[/MATH]
[MATH]\dfrac{-\ rc \pm \sqrt{c^2 \left ( r^2- \dfrac{4k}{c} \right)}}{2kc} =[/MATH]
[MATH]\dfrac{-\ r \cancel {c} \pm \cancel {c} \sqrt{r^2 - \dfrac{4k}{c}}}{2k \cancel {c}} =[/MATH]
[MATH]\dfrac{-\ r \pm \sqrt{r^2 - \dfrac{4k}{c}}}{2k} = [/MATH]
[MATH]\dfrac{-\ r \pm \sqrt{r^2 - \dfrac{4kr^2}{cr^2}}}{2k} = [/MATH]
[MATH]\dfrac{-\ r \pm \sqrt{r^2 \left (1 - \dfrac{4k}{cr^2} \right )}}{2k} =[/MATH]
[MATH]\dfrac{-\ r \pm r\sqrt{1 - \dfrac{4k}{cr^2}}}{2k} = \dfrac{-\ r}{2k} \pm \dfrac{r}{2k} \sqrt{1 - \dfrac{4k}{cr^2}}.[/MATH]

 

[dariel]

Excellent, thank you both for your time. I see the strategy now, in order to get the C and the R out.

I'm hoping this is a fixed form I will need to use, not sure I would have come up with that myself if presented with only the LHS.