Question

express in simplest radical form given $x > 0$.
$2x\sqrt{48x^3} + 9\sqrt{3x^5}$

Explanation:

Step1: Simplify each radical term

First, simplify \(2x\sqrt{48x^{3}}\). Factor \(48x^{3}\) as \(16x^{2}\cdot3x\) (since \(16x^{2}\) is a perfect square). Then \(\sqrt{48x^{3}}=\sqrt{16x^{2}\cdot3x}=\sqrt{16x^{2}}\cdot\sqrt{3x}=4x\sqrt{3x}\) (because \(x>0\), so \(\sqrt{x^{2}} = x\)). So \(2x\sqrt{48x^{3}}=2x\cdot4x\sqrt{3x}=8x^{2}\sqrt{3x}\).

Next, simplify \(9\sqrt{3x^{5}}\). Factor \(3x^{5}\) as \(x^{4}\cdot3x\) (since \(x^{4}\) is a perfect square). Then \(\sqrt{3x^{5}}=\sqrt{x^{4}\cdot3x}=\sqrt{x^{4}}\cdot\sqrt{3x}=x^{2}\sqrt{3x}\) (because \(x>0\), so \(\sqrt{x^{4}} = x^{2}\)). So \(9\sqrt{3x^{5}}=9\cdot x^{2}\sqrt{3x}=9x^{2}\sqrt{3x}\).

Step2: Combine like terms

Now we have \(8x^{2}\sqrt{3x}+9x^{2}\sqrt{3x}\). Since both terms have the same radical part \(x^{2}\sqrt{3x}\), we can combine the coefficients: \((8 + 9)x^{2}\sqrt{3x}=17x^{2}\sqrt{3x}\).

Answer:

\(17x^{2}\sqrt{3x}\)