How do you simplify #7sqrt(-36) + 4sqrt(-81)#?

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Answer 1 · 2015-11-15T11:19:18

0

#sqrt"-36"# = 0
#sqrt"-81"# = 0
The square root of a negative integer = 0
(7#times#0) + (4#times#0) = 0

Answer 2 · 2015-11-15T20:48:48

#78i#

If #x in RR and x>0#, then #sqrt(-x)=isqrtx#, where #i^2=-1 and i=sqrt(-1) in CC#.

Therefore,

#7sqrt(-36)+4sqrt(-81)=7*(6i)+4*(9i)#

#=42i + 36i#

#=78i#