How do you simplify # (sqrt(-4)+3)(2sqrt(-9)-1)#?

1 Answer
mason m
Mar 1, 2016

#-15+16i#

Explanation:

Recall the definition of #i#:

#i=sqrt(-1)#

This allows us to deal with square roots with negative numbers inside of them. The #2# in this problem are:

#sqrt(-4)=sqrt(2^2xx-1)=2i#

#sqrt(-9)=sqrt(3^2xx-1)=3i#

Substituting this into the original expression, we get:

#=(2i+3)(2(3i)-1)#

#=(2i+3)(6i-1)#

Now, distribute using the FOIL method.

#=underbrace(2i * 6i)_ "First"+underbrace(2i * -1)_ "Outside"+underbrace(3 * 6i)_ "Inside"+underbrace(3 * -1)_ "Last"#

#=12i^2-2i+18i-3#

#=12i^2+16i-3#

However, we are not done, since #i^2# can be simplified. Recall that since #i=sqrt(-1)#, we know that #i^2=-1#.

#=12(-1)+16i-3#

#=-15+16i#

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