Question

1. \\(sqrt{-11} cdot sqrt{-22}\\)

Explanation:

Step1: Rewrite using imaginary unit

Recall that \(\sqrt{-a}=i\sqrt{a}\) for \(a>0\). So, \(\sqrt{-11}=i\sqrt{11}\) and \(\sqrt{-22}=i\sqrt{22}\).

Step2: Multiply the expressions

Multiply the two rewritten expressions: \((i\sqrt{11})\cdot(i\sqrt{22})\). Using the property \(i\cdot i = i^{2}=- 1\) and \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), we get \(i^{2}\sqrt{11\times22}\).

Step3: Simplify the result

Substitute \(i^{2}=-1\) and simplify \(\sqrt{11\times22}\). First, \(11\times22 = 11\times11\times2=11^{2}\times2\), so \(\sqrt{11^{2}\times2}=11\sqrt{2}\). Then, \(i^{2}\sqrt{11\times22}=-1\times11\sqrt{2}=-11\sqrt{2}\).

Answer:

\(-11\sqrt{2}\)