Algebra Problem

[ku]

[math]w = \frac{K^{\frac{1}{4}}}{K^\frac{1}{2}}\frac{p}{2} \left(\frac{w}{2r}\right)^\frac{1}{2}[/math]

 

[math]r=\frac{1}{8}[/math] and [math]w = \frac{1}{2}[/math]

 

What is K in terms of p?

 

The answer is [math]K=4p^4[/math] but I want to see the working out because, for some reason, I got [math]K=2p^2[/math].

[ed84c]

looking at the two answers it looks as if you may have mixed up some surds somewhere.

[ed84c]

show us your working and it will be easier to help you out

[mcoy]

i think he's already got it.

if he didn't here's my working out.

 

[math]w = \frac{K^{\frac{1}{4}}}{K^\frac{1}{2}}\frac{p}{2} \left(\frac{w}{2r}\right)^\frac{1}{2}[/math]

 

[math]r=\frac{1}{8}[/math] and [math]w = \frac{1}{2}[/math]

 

= [math]\left[ w = \frac{K^{\frac{1}{4}}}{K^\frac{1}{2}}\frac{p}{2} \left(\frac{w}{2r}\right)^\frac{1}{2} \right]^4[/math]

 

= [math]w^4 = \frac{K}{K^2}\frac{p^4}{2^4}\left(\frac{w}{2r} \right)^2[/math]

 

or [math] w^4 = \frac{p^{4}w^{2}}{64Kr^2}[/math]

 

= [math] 64Kr^{2}w^{4}=p^{4}w^2 [/math]

 

then we make k the subject

 

[math] K = \frac{p^{4}w^2}{64r^{2}w^{4}}[/math]

or

[math] K = \frac{p^{4}}{64r^{2}w^2}[/math]

 

then lets make the substitution a little easier

 

[math] K = \left(\frac{p^{2}}{8rw}\right)^2[/math]

then substituting the e and w values we get

[math] K = \left(\frac{p^2}{1/2}\right)^2[/math]

[math] = [/math]

[math] K = (2p^2)^2 [/math]

[math] K = 4p^4 [/math]

 

i hope this helped.