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

Question

Elijah Allen · Accepted Answer

#-15+16i#

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#

Liam Brennan · Answer

undefined To simplify the expression $(\sqrt{-4}+3)(2\sqrt{-9}-1)$, we first simplify the square roots:

$\sqrt{-4} = 2i$ and $\sqrt{-9} = 3i$.

Now, we substitute these values back into the expression and perform the multiplication:

$(2i + 3)(2 \cdot 3i - 1)$

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

Next, we distribute:

$= 2i \cdot 6i - 2i \cdot 1 + 3 \cdot 6i - 3$

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

Since $i^2 = -1$, we have:

$= 12(-1) - 2i + 18i - 3$

$= -12 - 2i + 18i - 3$

Now, combine like terms:

$= -12 + 16i - 3$

$= -15 + 16i$

So, $(\sqrt{-4}+3)(2\sqrt{-9}-1)$ simplifies to $-15 + 16i$.