Changing the subject: solve r = 2q^2 / 3p for q=

[kien]

Is this the correct way to solve the question?

 

[stapel]

Is this the correct way to solve the question?

It is difficult to read your upside-down image. Does it contain the following?

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Make q the subject of the formula:

. . .\(\displaystyle r\, =\, \dfrac{2q^2}{3p}\)

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And is your work as follows?

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\(\displaystyle r\, =\, \dfrac{2q^2}{3p}\)

\(\displaystyle r\, (3p)\, =\, 2q^2\)

\(\displaystyle \dfrac{r\, (3p)}{2}\, =\, q^2\)

\(\displaystyle \dfrac{\sqrt{\strut r\, (3p)\,}}{2}\, =\, q\)

\(\displaystyle q\, =\, \dfrac{\sqrt{\strut r\, (3\,}\,p)}{2}\)

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If so, please explain your justification for taking only the positive square root of q², and please explain the contents of your radicals in the last two lines above. For instance, why is the denominator (the "2") not included, and what is going on in the last line? And why is " r (3p) " never simplified to "3pr"?

Please be complete (and right-side up). Thank you!

 

[Ishuda]

Is this the correct way to solve the question?

You think maybe you want to look at your post before you finalize it? It would be much easier on us if it weren't upside down which, at least to me, appears to be the case.