easy square root problem? medal given

Question

tester97 · Answer

attachment

tester97 · Answer

I think its B am i right?

tester97 · Answer

@Luigi0210 i just need you to seee if im correct

tester97 · Answer

actually i change my answer to C is this correct

anonymous · Answer

@wolfe8

wolfe8 · Answer

Here's a way to see it: you can rewrite this as$$(2401x ^{12}y ^{16})^{\frac{ 1 }{ 4 }}$$

tester97 · Answer

ok so now what?

wolfe8 · Answer

Now recall that when it's brackets you multiply the powers with 1/4. Remember do do it for 2401 too

wolfe8 · Answer

Aaaaand I'm trying to figure out why there are absolute brackets in there

anonymous · Answer

C is the right option @tester97 :)

tester97 · Answer

ahhh thanks thats what i got

wolfe8 · Answer

Actually, I think I know now. Since odd powers can give negative numbers, you will have to absolute it to get a positive whole result in the end.

opcode · Answer

$$\large (x^a)^b = x^{a\times b}$$
$$\sqrt[4]{2401x^{12}y^{16}}$$
Since the index is four:
$$\dfrac{12}{4}=3~and~\dfrac{16}{4}$$
$$\sqrt[4]{2401}x^{3}y^{3}$$
Now find a factor of $2401$ that is to the power of four:
$$2401 = 7^4$$
$$\sqrt[4]{2401x^{12}y^{16}} = 7|x^3|y^4$$
$$|~and~|~\text{are used to show that it is not}~7x^3~but~7~x^3$$

opcode · Answer

Oops I meant $y^4$

tester97 · Answer

So i was right?

opcode · Answer

Yes, I just wanted to show how to get it if anyone was interested :-P.

tester97 · Answer

oh ok thanks :)