Question

Please help, medal and fan

Answer 1

Assuming x in nonnegative,
\(\sqrt{x^2} = x\)

Answer 2

\(x^6 = x^2 \times x^2 \times x^2\)

Answer 3

true, so \[x ^{6} = x ^{3} ?\]

Answer 4

I think you meant \(\sqrt{x^6} = x^3\)
which is correct.

Answer 5

yeah, thats what i meant. but then we have that 24 attached?

Answer 6

That takes care of the x part.
Now we need to take care of the number part.

Answer 7

Can you factor 24 into a product of two numbers, one of which being a perfect square?

Answer 8

since 4 is a perfect square we could split it up into \[\sqrt{4 * 6}\]

Answer 9

and then, keep simplifying and we get \[2\sqrt{6}\] , right?

Answer 10

Exactly.
Now split the root into a product of roots, and take the root of 4.
\(\sqrt{24} = \sqrt{4 \times6} = \sqrt 4 \sqrt6\)

Answer 11

Correct.

Answer 12

so its c?

Answer 13

Now put it all together, the x part and the number part.

Answer 14

Correct.

Answer 15

Thanks for your help! :D

Answer 16

You're welcome.

Answer 17

The general idea here is to split up the radicand into a product of squares and what's left over.
Then every square can have its square root taken. What is left stays inside the root.

Answer 18

ok, i understand. :D

Answer 19

Great.
You're welcome.