Simplifying Radicals with varibles?

[liz777]

I understand that the square root of (for example)j^16 is j^8. But when you have an odd squared root, like the square root of j^19, would it be j^9square root of j?

Another quick question 12 radical 36 would be 72?

Sorry, I know this is basic algebra, but I really have forgotten how to do it...

 

[Mentallic]

Delete - Multipost.

 

[Mentallic]

Yes that is exactly what it would be.

$$\sqrt{x}=x^{\frac{1}{2}}$$
and
$$\sqrt{x^a}=x^{\frac{a}{2}$$

So for your question $$\sqrt{j^{16}}=j^{\frac{16}{2}}=j^8$$

Thus, $$\sqrt{j^19}=j^{\frac{19}{2}}=j^{9\frac{1}{2}}=j^9\sqrt{j}$$


For the second one, $$12\sqrt{36}=72$$ because $$\sqrt{36}=6$$ therefore $$12x6=72$$

If there's something you still don't understand, just ask

 

[naresh]

I would add a plus or minus to every result

 

[Mentallic]

Well if you want to be wrong then go ahead.