Question

1. $2sqrt{128}$

Explanation:

Step1: Factor 128

We know that \(128 = 64\times2\), and \(64\) is a perfect square (\(8^2 = 64\)). So we can rewrite \(\sqrt{128}\) as \(\sqrt{64\times2}\).
\[
\sqrt{128}=\sqrt{64\times2}
\]

Step2: Use square - root property

According to the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (\(a\geq0,b\geq0\)), we have \(\sqrt{64\times2}=\sqrt{64}\times\sqrt{2}\). Since \(\sqrt{64} = 8\), then \(\sqrt{128}=8\sqrt{2}\).
\[
\sqrt{128}=8\sqrt{2}
\]

Step3: Multiply by 2

Now we have the original expression \(2\sqrt{128}\), substitute \(\sqrt{128}=8\sqrt{2}\) into it, we get \(2\times8\sqrt{2}\).
\[
2\sqrt{128}=2\times8\sqrt{2}
\]

Step4: Calculate the product

Calculate \(2\times8 = 16\), so \(2\sqrt{128}=16\sqrt{2}\).
\[
2\times8\sqrt{2}=16\sqrt{2}
\]

Answer:

\(16\sqrt{2}\)