How do you simplify #2sqrt(-49) + 3sqrt(-64)#?

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Answer 1 · 2015-11-17T12:22:23

#=+-38i or +-10i#

From complex number theory, #i=sqrt(-1) and i^2=-1#.

Note also that #(ai)^2=(-ai)^2=-1# and hence #sqrt(-a)=+-isqrta# , #AAa in RR^+#.

#therefore2sqrt(-49)+3sqrt(-64)=2*(+-7i)+3*(+-8i)#

#=+-14i+-24i#

#=+-38i or +-10i#

Answer 2 · 2015-11-17T12:22:49

#34i#

Remember that #i^2 = 1# so #sqrt(-1) = i#.

Keeping this in mind, you can simplify the term as follows:

#color(white)(xx) 2 sqrt(-49) + 3 sqrt(-64)#

#= 2 * sqrt(-1) * sqrt(49) + 3 * sqrt(-1) * sqrt(64)#

#= 2 * i * 7 + 3 * i * 8 #

#= 14i + 24i#

#= 34i#