When simplified, what is the value if i2 = −1?

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Question 2

Solution:

Given:

$\begin{gathered} \sqrt[]{-128} \\ \\ \text{also,} \\ i^2=-1 \end{gathered}$

To simplify, we break the radical down into bits.

$\begin{gathered} \sqrt[]{-128}=\sqrt[]{-1}*\sqrt[]{128} \\ To\text{ simplify }\sqrt[]{128}\text{ , we split it into a number that is a p}\operatorname{erf}ect\text{ square and another number,} \\ \text{Hence,} \\ \sqrt[]{128}=\sqrt[]{64*2}=\sqrt[]{64}*\sqrt[]{2} \\ \text{This means,} \\ \sqrt[]{-128}=\sqrt[]{-1}*\sqrt[]{64}*\sqrt[]{2} \\ \text{But recall that;} \\ i^2=-1 \\ \text{Taking the square root of both sides,} \\ i=\sqrt[]{-1} \\ \text{Substituting it in the expression below,} \\ \sqrt[]{-128}=\sqrt[]{-1}*\sqrt[]{64}*\sqrt[]{2} \\ \sqrt[]{-128}=i*8*\sqrt[]{2} \\ \sqrt[]{-128}=8i\sqrt[]{2} \end{gathered}$

Therefore, the simplified form is;

$8i\sqrt[]{2}$

When simplified, what is the value if i2 = −1?

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