√-128 in simplest radicle form

1 answer

To simplify \(\sqrt{-128}\), we can first rewrite it using the imaginary unit \(i\), which is defined as \(i = \sqrt{-1}\).

  1. We start with \(\sqrt{-128} = \sqrt{128} \cdot \sqrt{-1} = \sqrt{128} \cdot i\).

  2. Next, we simplify \(\sqrt{128}\). We can factor \(128\) into its prime factors:

    \[ 128 = 64 \times 2 = 8^2 \times 2 = (4^2 \times 4) \times 2 = (2^6 \times 2) = 2^7 \]

  3. To find \(\sqrt{128}\), we can take the square root of the perfect square factor:

    \[ \sqrt{128} = \sqrt{64 \times 2} = \sqrt{64} \cdot \sqrt{2} = 8\sqrt{2} \]

  4. Now, we can substitute this back into our expression for \(\sqrt{-128}\):

    \[ \sqrt{-128} = 8\sqrt{2} \cdot i \]

Thus, the simplest radical form of \(\sqrt{-128}\) is

\[ 8\sqrt{2}i \]