Simplify sqrt(48x^12) / sqrt(2x) using the quotient rule for square roots, assuming x > 0.

Step-by-step solution

1. 1

   Merge the two radicals before simplifying either one. The quotient rule reads

   $\frac{\sqrt{A}}{\sqrt{B}}=\sqrt{\frac{A}{B}}\quad (A\ge 0,\ B>0),$

   and it is far less work here than simplifying $\sqrt{48x^{12}}$ and $\sqrt{2x}$ separately. The assumption $x>0$ is what guarantees both radicands are non-negative and the denominator is non-zero.

2. 2

   Divide inside the single radical. Handle the number and the powers separately:

   $\frac{48}{2}=24,\qquad \frac{x^{12}}{x^{1}}=x^{11}\ \Longrightarrow\ \sqrt{\frac{48x^{12}}{2x}}=\sqrt{24x^{11}}.$

3. 3

   Split off the largest perfect squares. For the coefficient, $24=4\cdot 6$ with $4=2^2$. For the variable, the largest even power below $11$ is $10$, so $x^{11}=x^{10}\cdot x$ with $x^{10}=(x^5)^2$:

   $\sqrt{24x^{11}}=\sqrt{4\cdot 6\cdot x^{10}\cdot x}=\sqrt{4}\cdot\sqrt{x^{10}}\cdot\sqrt{6x}.$

4. 4

   Take the roots that come out. Because $x>0$, no absolute-value bars are needed:

   $2\cdot x^{5}\cdot\sqrt{6x}=2x^{5}\sqrt{6x}.$

5. 5

   Confirm the result is fully simplified and correct. Under the remaining radical, $6=2\cdot 3$ is square-free and $x$ appears to the first power, so nothing more escapes. Numerically at $x=2$: $\sqrt{48\cdot 4096}/\sqrt{4}=\sqrt{196608}/2=221.7025$, and $2(32)\sqrt{12}=64(3.4641)=221.7025$. The two agree.