![\bf \cfrac{\sqrt[4]{63}}{4\sqrt[4]{6}}\qquad \begin{cases} 63=3\cdot 3\cdot 7\\ 6=2\cdot 3 \end{cases}\implies \cfrac{\sqrt[4]{3\cdot 3\cdot 7}}{4\sqrt[4]{2\cdot 3}}\implies \cfrac{\underline{\sqrt[4]{3}}\cdot \sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}\cdot \underline{\sqrt[4]{3}}} \\\\\\ \cfrac{\sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{3\cdot 7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B63%7D%7D%7B4%5Csqrt%5B4%5D%7B6%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A63%3D3%5Ccdot%203%5Ccdot%207%5C%5C%0A6%3D2%5Ccdot%203%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%203%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%203%7D%7D%5Cimplies%20%5Ccfrac%7B%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%5Ccdot%20%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D)
![\bf \textit{now, rationalizing the denominator}\\\\ \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}\cdot \cfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21}\cdot \sqrt[4]{8}}{4\sqrt[4]{2}\cdot \sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21\cdot 8}}{4\sqrt[4]{2\cdot 2^3}}\implies \cfrac{\sqrt[4]{168}}{4\sqrt[4]{2^4}} \\\\\\ \cfrac{\sqrt[4]{168}}{4\cdot 2}\implies \cfrac{\sqrt[4]{168}}{8}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bnow%2C%20rationalizing%20the%20denominator%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%5Ccdot%20%5Csqrt%5B4%5D%7B8%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%5Ccdot%208%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%202%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5E4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Ccdot%202%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B8%7D)
and is all you can simplify from it.
so... all we did, was rationaliize it, namely, "getting rid of the pesky radical at the bottom", we do so by simply multiplying it by something that will raise the radicand, to the same degree as the root, thus the radicand comes out.
Whatever you do to one side, you have to do to the other.
Try to get all the x values and regular values together.
To do that, just subtract both sides by 6x to cancel it out on one side and subtract it from the other side:
6x-6x+32=-2x-6x
=
32=-8x
now divide both sides by -8 to get the x by itself
32/-8=-8/-8x
= 32/-8=x
-4=x
so your answer is: x= -4
Is there not more to this problem? maybe thats the final answers
Assuming I understand your question correctly, in that you’re looking for just some descriptions of the differences between the functions. If so, then I’d say:
First graph both functions, the f(x) and the g(x). Then spot the differences.
Note that the g(x) function has shifted towards the right compared with the f(x) function.
Another way that the g(x) differs from the f(x) function is that it’s stretched. The vertex is in the IV quadrant for g(x) rather than at the origin for f(x).
I hope that helps.
