## anonymous 5 years ago hello, can you explain : 3 square root x^6 divided 3 square root 8x multiplied by 3 square root x^11. ???

Hm. Let me see if I have this right: $\frac{3\sqrt{x^6}}{3\sqrt{8x}} \cdot 3\sqrt{x^{11}}$ Is that what the problem looks like?

2. anonymous

the 3 sqroot x^11 is under the radicand.

Ok, gotcha. So we have: $\frac{3\sqrt{x^6}}{3\sqrt{8x}\cdot 3\sqrt{x^{11}}}$

We can combine like terms by multiplying the 3s together and moving both of the bottom roots into the same one: $\frac{3\sqrt{x^6}}{9\sqrt{8x\cdot x^{11}}}$ We see that we can combine the x terms in the denominator: $\frac{3\sqrt{x^6}}{9\sqrt{8x^{12}}}$

5. anonymous

thank you so very much!

Next, we can actually take the roots. Since we have even powers, that's easy: $\frac{3x^3}{9x^6\sqrt{8}}$ Then we can divide properly for the powers and the 3/9: $\frac{1}{3x^3\sqrt{8}}$

Finally, we can simplify $$\sqrt{8}$$ by remembering that 8 is 4 * 2 and that we can take the square root of 4: $\frac{1}{3x^3(2\sqrt{2})}$ $\frac{1}{6x^3\sqrt{2}}$

Most people also prefer keeping radicals out of the denominator; to do that, we can multiply by $$\sqrt{2}/\sqrt{2}$$ and move the radical to the numerator: $\frac{1}{6x^3\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}$ $\frac{\sqrt{2}}{6x^3\cdot 2}$ $\frac{\sqrt{2}}{12x^3}$

No problem, glad to help :)

10. anonymous

thanks!