Question

Solve, in standard form.2 sqrt(-8) x 3sqrt(-48)

Answer 1

`2\sqrt[]{-8}\cdot3√(-48)`

To solve the given expression, we can use the following property of radicals:

`\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\Rightarrow\text{ Product property}`

Then, we have:

`\begin{gathered} 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=2\cdot3\sqrt[]{-8}\cdot\sqrt[]{-48} \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8}\sqrt[]{-1}\cdot\sqrt[]{48}\sqrt[]{-1} \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8}i\cdot\sqrt[]{48}i \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8}\cdot\sqrt[]{48}i\cdot i \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8}\sqrt[]{48}i^2 \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8\cdot48}i^2 \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8\cdot8\cdot6}i^2 \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8^2\cdot6}i^2 \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\sqrt[]{8^2}\cdot\sqrt[]{6}i^2 \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=6\cdot8\cdot\sqrt[]{6}i^2 \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=48\sqrt[]{6}i^2 \\ 2\sqrt[]{-8}\cdot3\sqrt[]{-48}=-48\sqrt[]{6}\Rightarrow\text{ Because }i^2=-1 \end{gathered}`

Therefore, the solution of the given expression is:

`$$\boldsymbol{-48\sqrt[]{6}}$$`