I need help with this question

[Holdstill123]

Help please

 

[Deleted member 4993]

Help please

Please follow the rules of posting in this forum, as enunciated at:

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Please share your work/thoughts about this assignment.

 

[JeffM]

Please do as Subhotosh Khan has asked.

While thinking about the problem, it may help to remember these four fundamental rules about roots:

[MATH]\text {Given positive real numbers } a,\ b,\ c \text { and } d \text { and integer } n \ge 2:[/MATH]
[MATH]\sqrt{a} \equiv \sqrt[2]{a};[/MATH]
[MATH]b = \sqrt[n]{a} \iff b^ n = a;[/MATH]
[MATH]\sqrt[n]{c * d} \equiv \sqrt[n]{c} * \sqrt[n]{d} \text { and }[/MATH]
[MATH]\sqrt[n]{\dfrac{c}{d}} \equiv \dfrac{\sqrt[n]{c}}{\sqrt[n]{d}}.[/MATH]
So for example

[MATH]4^2 = 16 \implies 4 = \sqrt[2]{16} = \sqrt{16}.[/MATH]

 

[Steven G]

I do not think this problem can be done as stated.

 

[pka]

I do not think this problem can be done as stated.

I basically agree the above and suspect this question was written by a rank amateur.
If stipulate that \(\displaystyle x>0\) then write it as \(\displaystyle \sqrt{\dfrac{16a}{c^6x^2}}\)

 

[Steven G]

I basically agree the above and suspect this question was written by a rank amateur.
If stipulate that \(\displaystyle x>0\) then write it as \(\displaystyle \sqrt{\dfrac{16a}{c^6x^2}}\)

Why not c>0 as well?

 

[pka]

Why not c>0 as well?

Because from the given we must assume that \(\displaystyle c^3\cdot x\ne 0\).
Thus we know that \(\displaystyle c^6>0\) Now that I think about it may need \(\displaystyle c>0\).