Question

think now: 12.15.2025
add and/or subtract the given.
\sqrt{96x^2} + \sqrt{24x^2}
\bigcirc 6x\sqrt{6x}
\bigcirc 8x\sqrt{6}
\bigcirc 6x\sqrt{6}
\bigcirc 8x\sqrt{6x}

Explanation:

Step1: Simplify each square root

For \(\sqrt{96x^2}\), factor 96 as \(16\times6\) and \(x^2\) is a perfect square. So \(\sqrt{96x^2}=\sqrt{16\times6\times x^2}=\sqrt{16}\times\sqrt{x^2}\times\sqrt{6}=4x\sqrt{6}\) (assuming \(x\geq0\) for real - valued square roots).
For \(\sqrt{24x^2}\), factor 24 as \(4\times6\) and \(x^2\) is a perfect square. So \(\sqrt{24x^2}=\sqrt{4\times6\times x^2}=\sqrt{4}\times\sqrt{x^2}\times\sqrt{6}=2x\sqrt{6}\).

Step2: Add the simplified square roots

Now we add \(4x\sqrt{6}+2x\sqrt{6}\). Since the terms have the same radical part (\(\sqrt{6}\)) and the same variable part (\(x\)), we can combine the coefficients: \((4x + 2x)\sqrt{6}=6x\sqrt{6}\).

Answer:

\(6x\sqrt{6}\) (corresponding to the option " \(6x\sqrt{6}\)")