Question

can some help with roots of real number.
1.√ (2x)^8

Answer 1

Simplify\[(\sqrt{2x})^{2}(\sqrt{2x})^{2}(\sqrt{2x})^{2}\]

Answer 2

the idea with simplifying square roots is to get a pair of "the same thing"
then you can rewrite
\[ \sqrt{a \cdot a} \text{ as } a\]
(you take away the square root, and write only 1 of the pair.

For example, if you had
\[ \sqrt{3^4} \]
3^4 means 3*3*3*3 (3 times itself 4 times)
\[ \sqrt{3 \cdot 3 \cdot 3 \cdot 3} \]
you can see there are two pairs of 3's (3*3) and (3*3)
you can rewrite this as
\[ \sqrt{3 \cdot 3 \cdot 3 \cdot 3} =3 \cdot 3\]
or if we multiplied everything out , using 3^4= 3*3*3*3=81 and 3*3=9,
\[ \sqrt{81} =9 \]

Answer 3

For your problem, (2x)^8 is (2x) times itself 8 times. Can you write this as pairs of (2x) ?