Question

Please Help Can someone explain where the 4i to the square root of 3 came from?

Answer 1

Answer:

see explanation

Step-by-step explanation:

note that $\sqrt{-1}$ = i

$\sqrt{64-112}$

= $\sqrt{-48}$

= $\sqrt{16(3)(-1)}$

= $\sqrt{16}$ × $\sqrt{3}$ × $\sqrt{-1}$

= 4 × $\sqrt{3}$ × i

= 4i$\sqrt{3}$