▶️ Ejercicio 253 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXII

CANTIDADES IMAGINARIAS

Reducir a la forma de una cantidad real multiplicada por

\sqrt{- 1} o i

$\begin{matrix} \sqrt{- a^{2}} & = \sqrt{(- 1) (a^{2})} \\ = \sqrt{- 1} a \\ = a i \end{matrix}$
$\begin{matrix} \sqrt{- 2} & = \sqrt{(- 1) (2)} \\ = \sqrt{- 1} \sqrt{2} \\ = \sqrt{2} i \end{matrix}$
$\begin{matrix} 2 \sqrt{- 9} & = 2 \sqrt{- 3^{2}} \\ = 2 \sqrt{(- 1) {(3)}^{2}} \\ = 2.3 \sqrt{- 1} \\ = 6 i \end{matrix}$
$\begin{matrix} \sqrt{- 81} & = \sqrt{- 9^{2}} \\ = 3 \sqrt{- 1} \\ = 3 i \end{matrix}$
$\begin{matrix} \sqrt{- 6} & = \sqrt{(- 1) (6)} \\ = \sqrt{- 1} \sqrt{6} \\ = \sqrt{6} i \end{matrix}$
$\begin{matrix} 3 \sqrt{- b^{4}} & = 3 \sqrt{(- 1) (b^{4})} \\ = 3 b^{2} \sqrt{- 1} \\ = 3 b^{2} i \end{matrix}$
$\begin{matrix} \sqrt{- 12} & = \sqrt{(- 1) (2^{2}) (3)} \\ = 2 \sqrt{- 1} \sqrt{3} \\ = 2 \sqrt{3} i \end{matrix}$
$\begin{matrix} \sqrt{- 7} & = \sqrt{(- 1) (7)} \\ = \sqrt{- 1} \sqrt{7} \\ = \sqrt{7} i \end{matrix}$
$\begin{matrix} \sqrt{- 27} & = \sqrt{(- 1) (3^{2}) (3)} \\ = 3 \sqrt{- 1} \sqrt{3} \\ = 3 \sqrt{3} i \end{matrix}$
$\begin{matrix} \sqrt{- 4 m^{4}} & = \sqrt{(- 1) (2^{2}) (m^{4})} \\ = 2 m^{2} \sqrt{- 1} \\ = 2 m^{2} i \end{matrix}$
$\begin{matrix} \sqrt{- \frac{1}{16}} & = \sqrt{(- 1) (\frac{1}{4^{2}})} \\ = \frac{1}{4} \sqrt{- 1} \\ = \frac{1}{4} i \end{matrix}$
$\begin{matrix} \sqrt{- a^{2} - b^{2}} & = \sqrt{- (a^{2} + b^{2})} \\ = \sqrt{(- 1) (a^{2} + b^{2})} \\ = \sqrt{- 1} \sqrt{a^{2} + b^{2}} \\ = \sqrt{a^{2} + b^{2}} i \end{matrix}$