Question

lets do some more problems with higher order roots
problem 3.1
$sqrt{128} = square$
check
explain

Explanation:

Step1: Factor 128

We know that \(128 = 64\times2\), and \(64 = 8^2\) (or \(64 = 2^6\), but for square root, we look for the largest perfect square factor). So we can write \(\sqrt{128}=\sqrt{64\times2}\).

Step2: Use square root property

Using the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (where \(a\geq0\) and \(b\geq0\)), we have \(\sqrt{64\times2}=\sqrt{64}\times\sqrt{2}\).
Since \(\sqrt{64} = 8\) (because \(8\times8 = 64\)), then \(\sqrt{64}\times\sqrt{2}=8\sqrt{2}\).

Answer:

\(8\sqrt{2}\)