Question

Show all work to multiply quantity $6$ plus the square root of negative $64$ end quantity times quantity $3$ minus the square root of negative $16$ end quantity.

Answer 1

Step 1: Convert the word sentence to numeric expression.

The square root of negative $64$ is $\sqrt{-64}$ and square root of negative $16$ is $\sqrt{-16}$.

$$\begin{gathered} \therefore \left(6+\sqrt{-64}\right)\left(3-\sqrt{-16}\right) \\ =\left(6+\sqrt{-1}\times \sqrt{64}\right)\left(3-\sqrt{-1}\times \sqrt{16}\right) \end{gathered}$$

Step 2: Simplify the expression $\left(6+\sqrt{-1}\times \sqrt{64}\right)\left(3-\sqrt{-1}\times \sqrt{16}\right)$.

Substitute the value of $\sqrt{-1}$as $i.$

Multiply the obtained expression.

$\begin{array}{rcl}\left(6+\sqrt{-1}\times \sqrt{64}\right)\left(3-\sqrt{-1}\times \sqrt{16}\right)& =& \left(6+\sqrt{64}i\right)\left(3-\sqrt{16}i\right)\\ & =& \left(6+8i\right)\left(3-4i\right)\\ & =& 18-24i+24i-32i^2\end{array}$

Step 3: Simplify the expression $18-24i+24i-32i^2$.

Substitute the value $-1$ for $i^2$ and then simplify.

$\begin{array}{rcl}18-24i+24i-32i^2& =& 18-32\left(-1\right)\\ & =& 18+32\\ & =& 50\end{array}$

Therefore, the value of the given equation is $50.$