## Algebra 2 Common Core

$10-4i$
Since $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b},$ the given expression, $(4+\sqrt{-9})+(6-\sqrt{-49}) ,$ is equivalent to \begin{align*} & (4+\sqrt{-1}\cdot\sqrt{9})+(6-\sqrt{-1}\cdot\sqrt{49}) \\&= (4+\sqrt{-1}\cdot3)+(6-\sqrt{-1}\cdot7) \\&= (4+3\sqrt{-1})+(6-7\sqrt{-1}) .\end{align*} Since $i=\sqrt{-1},$ the expression above is equivalent to \begin{align*} (4+3i)+(6-7i) .\end{align*} Removing the grouping symbols, the expression above is equivalent to \begin{align*} 4+3i+6-7i .\end{align*} Combining the real parts and the imaginary parts, the expression above is equivalent to \begin{align*} & (4+6)+(3i-7i) \\&= 10+(-4i) \\&= 10-4i .\end{align*} Hence, the simplified form of the given expression is $10-4i$.