Question

simplify using complex numbers: \\(\sqrt{-108x^2}\\)

Explanation:

Step1: Recall the imaginary unit

Recall that \( \sqrt{-a}=\sqrt{a}\cdot\sqrt{-1}=i\sqrt{a} \) for \( a\geq0 \), where \( i \) is the imaginary unit with \( i^{2}=-1 \). So we can rewrite \( \sqrt{- 108x^{2}} \) as \( \sqrt{108x^{2}}\cdot\sqrt{-1} \).

Step2: Simplify the square - root of the real part

Simplify \( \sqrt{108x^{2}} \). First, factorize \( 108 \): \( 108 = 36\times3=6^{2}\times3 \), and \( x^{2}=(x)^{2} \) (assuming \( x\in\mathbb{R} \)). Then \( \sqrt{108x^{2}}=\sqrt{6^{2}\times3\times x^{2}} \). Using the property \( \sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \) (\( a\geq0,b\geq0 \)) and \( \sqrt{a^{2}} = |a| \), we have \( \sqrt{6^{2}\times3\times x^{2}}=\sqrt{6^{2}}\cdot\sqrt{x^{2}}\cdot\sqrt{3}=6|x|\sqrt{3} \). But if we assume \( x \) is a real number and we are simplifying in the context of complex numbers (usually we can consider \( x\geq0 \) for the purpose of simplifying the square - root of \( x^{2} \) as \( x \), unless specified otherwise), so \( \sqrt{x^{2}}=x \) (for \( x\geq0 \)). So \( \sqrt{108x^{2}} = 6x\sqrt{3} \) (for \( x\geq0 \)).

Step3: Combine with the imaginary unit

Since \( \sqrt{-108x^{2}}=\sqrt{108x^{2}}\cdot\sqrt{-1} \), substituting the simplified form of \( \sqrt{108x^{2}} \) and \( \sqrt{-1}=i \), we get \( \sqrt{-108x^{2}}=6x\sqrt{3}\cdot i = 6i\sqrt{3}x \) (or \( 6\sqrt{3}xi \)).

Answer:

\( 6\sqrt{3}xi \) (or \( 6i\sqrt{3}x \))