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Simplifying Radicals
What is the simplified form of
sqrt(48n^(9)) ?
(1 point)

4n^(3)sqrt3

4n^(4)sqrt(3n)

3nsqrt(4n^(8))

4sqrt(3n^(9))

Simplifying Radicals $\newline$ What is the simplified form of $\sqrt{48 n^{9}}$ ? $\newline$ ( $1$ point) $\newline$ $4 n^{3} \sqrt{3}$ $\newline$ $4 n^{4} \sqrt{3 n}$ $\newline$ $3 n \sqrt{4 n^{8}}$ $\newline$ $4 \sqrt{3 n^{9}}$

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Simplify radical expressions with root inside the root

Full solution

Q. Simplifying Radicals

\newline

What is the simplified form of

\sqrt{48 n^{9}}

?

\newline

(

1

point)

\newline

4 n^{3} \sqrt{3}

\newline

4 n^{4} \sqrt{3 n}

\newline

3 n \sqrt{4 n^{8}}

\newline

4 \sqrt{3 n^{9}}

Factor $48$ and $n^9$ : First, we need to factor $48$ into its prime factors and express $n^9$ in terms of squares to simplify the square root. $\newline$ $48$ can be factored into $2^4 \times 3$ (since $48 = 16 \times 3$ and $16 = 2^4$ ). $\newline$ $n^9$ can be expressed as $(n^4)^2 \times n$ , since $n^9 = n^8 \times n$ and $n^8$ is a perfect square.
Rewrite $\sqrt{48n^{9}}$ : Now, we can rewrite $\sqrt{48n^{9}}$ using the factors from the previous step. $\newline$ $\sqrt{48n^{9}} = \sqrt{2^4 \times 3 \times (n^4)^2 \times n}$
Take out squares: Next, we can take out the squares from under the square root. $\newline$ $\sqrt{2^4 \times 3 \times (n^4)^2 \times n} = \sqrt{2^4} \times \sqrt{(n^4)^2} \times \sqrt{3n}$ $\newline$ $= 2^2 \times n^4 \times \sqrt{3n}$ $\newline$ $= 4n^4 \times \sqrt{3n}$
Simplify radical: We have simplified the radical to its simplest form.

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