Question

perform the indicated operations and write the result in standard form.
\\((4\sqrt{-12})(-6\sqrt{-12})\\)

\\((4\sqrt{-12})(-6\sqrt{-12}) = \square\\)
(type an exact answer, using radicals as needed. type your answer in the form \\(a + bi\\).)

Explanation:

Step1: Recall the imaginary unit \(i = \sqrt{-1}\)

Rewrite \(\sqrt{-12}\) as \(\sqrt{12}\cdot\sqrt{-1}=2\sqrt{3}i\)

Step2: Substitute \(\sqrt{-12}\) with \(2\sqrt{3}i\) in the expression

The expression \((4\sqrt{-12})(-6\sqrt{-12})\) becomes \((4\cdot2\sqrt{3}i)(-6\cdot2\sqrt{3}i)\)

Step3: Multiply the coefficients and the imaginary units

First, multiply the coefficients: \(4\times2\times(-6)\times2 = -96\)
Then, multiply the radicals: \(\sqrt{3}\times\sqrt{3}=3\)
And multiply the imaginary units: \(i\times i = i^{2}=-1\)

Step4: Combine the results

Multiply all the parts together: \(-96\times3\times(-1)=288\) (Wait, let's do it step by step. Wait, actually, let's re - do the multiplication correctly.

Original expression after substituting \(i\):

\((4\sqrt{-12})(-6\sqrt{-12})=4\times(-6)\times\sqrt{-12}\times\sqrt{-12}\)

We know that \(\sqrt{-12}\times\sqrt{-12}=(\sqrt{-12})^{2}=- 12\) (because \((\sqrt{a})^{2}=a\) for \(a\geq0\), but here \(a = - 12\), and using \(i^{2}=-1\), \((\sqrt{-12})^{2}=(-12)\))

So \(4\times(-6)\times(-12)=4\times72 = 288\)

Wait, another way:

\(\sqrt{-12}=\sqrt{12}i = 2\sqrt{3}i\)

So \((4\times2\sqrt{3}i)\times(-6\times2\sqrt{3}i)=(8\sqrt{3}i)\times(-12\sqrt{3}i)\)

Multiply the coefficients: \(8\times(-12)=-96\)

Multiply the radicals: \(\sqrt{3}\times\sqrt{3}=3\)

Multiply the imaginary units: \(i\times i=i^{2}=-1\)

Then \(-96\times3\times(-1)=288\) (because \(-96\times3=-288\) and \(-288\times(-1) = 288\))

Or, using the property \(\sqrt{-a}\times\sqrt{-b}=-\sqrt{ab}\) (when \(a,b>0\))

\(\sqrt{-12}\times\sqrt{-12}=-\sqrt{12\times12}=- 12\)

Then \(4\times(-6)\times(-12)=4\times72 = 288\)

Answer:

\(288 + 0i\) (or simply \(288\))