1. ## Having difficulty understanding radicals

Will this cause me massive problems in the future if I don't fully 100% understand radicals? For example:

Simplify $\displaystyle \sqrt[4]{\frac{3a}{2b^3c^2}}$

I just don't know where to begin... I'm looking in the book for the answer and it says this.

$\displaystyle \sqrt[4]{\frac{3a}{2b^3c^2}}=\sqrt[4]{\frac{3a}{2b^3c^2}\cdot\frac{8bc^2}{8bc^2}}$ why $\displaystyle 8bc^2$?

Then the next step: $\displaystyle \sqrt[4]{\frac{24abc^2}{16b^4c^4}}$ Which I understand obviously, but then the final step is $\displaystyle \frac{1}{2bc}\sqrt[4]{24abc^2}$

This is super complicated to me and it only gets worse... do I really need to know this?

2. ## Re: Having difficulty understanding radicals

Originally Posted by uperkurk
Simplify $\displaystyle \sqrt[4]{\frac{3a}{2b^3c^2}}$
$\displaystyle \sqrt[4]{\frac{3a}{2b^3c^2}}=\sqrt[4]{\frac{3a}{2b^3c^2}\cdot\frac{8bc^2}{8bc^2}}$ why $\displaystyle \color{red}8bc^2$?
Because you need a perfect $\displaystyle 4^{th}$ power in that denominator: $\displaystyle 2^4b^4c^4$.

3. ## Re: Having difficulty understanding radicals

Originally Posted by uperkurk
Will this cause me massive problems in the future if I don't fully 100% understand radicals? For example:

Simplify $\displaystyle \sqrt[4]{\frac{3a}{2b^3c^2}}$

I just don't know where to begin... I'm looking in the book for the answer and it says this.

$\displaystyle \sqrt[4]{\frac{3a}{2b^3c^2}}=\sqrt[4]{\frac{3a}{2b^3c^2}\cdot\frac{8bc^2}{8bc^2}}$ why $\displaystyle 8bc^2$?

Then the next step: $\displaystyle \sqrt[4]{\frac{24abc^2}{16b^4c^4}}$ Which I understand obviously, but then the final step is $\displaystyle \frac{1}{2bc}\sqrt[4]{24abc^2}$

This is super complicated to me and it only gets worse... do I really need to know this?
$\displaystyle 2^4= 16$. Since, in the original problem, there was a "2" in the denominator, they multiplied by 8 to get a "perfect fourth power". Similarly, $\displaystyle b^3(b)= b^4$ and $\displaystyle c^2(c^2)= c^4$. Multiplying the denominator by $\displaystyle 8bc^2$ makes the denominator $\displaystyle 16b^4c^4= (2bc)^4$ so that $\displaystyle \sqrt[4]{(2bc)^4}= 2bc$.

Of course, if you multiply by $\displaystyle 8bc^2$ in the denominator you must do the same in the numerator.

4. ## Re: Having difficulty understanding radicals

Originally Posted by HallsofIvy
$\displaystyle 2^4= 16$. Since, in the original problem, there was a "2" in the denominator, they multiplied by 8 to get a "perfect fourth power". Similarly, $\displaystyle b^3(b)= b^4$ and $\displaystyle c^2(c^2)= c^4$. Multiplying the denominator by $\displaystyle 8bc^2$ makes the denominator $\displaystyle 16b^4c^4= (2bc)^4$ so that $\displaystyle \sqrt[4]{(2bc)^4}= 2bc$.

Of course, if you multiply by $\displaystyle 8bc^2$ in the denominator you must do the same in the numerator.
Ok I see, but what if it was $\displaystyle 3b^2c^4$ for example?

5. ## Re: Having difficulty understanding radicals

Originally Posted by uperkurk
Ok I see, but what if it was $\displaystyle 3b^2c^4$ for example?
Then use $\displaystyle 27b^2$ to get $\displaystyle 3^4b^4c^4$.

6. ## Re: Having difficulty understanding radicals

Oh right so you basically $\displaystyle 3b^2c^4\cdot 27b^2 = 81b^4c^4$ but then notice that 81 is $\displaystyle 3^4$ ok I see. It's pretty difficult because you need to think of a number that when multiplied, the 4th root will the original number, in our case 3.