How do you simplify #(7sqrt(-3))(2sqrt(-27))#?

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Sarah Share
Mar 8, 2018

Answer:

Always deal with imaginary numbers first. Final answer: #-126#

Explanation:

If you ignore the rule of imaginary numbers which states that #sqrt(-1)=##i# and #i^2#=-1, you would be tempted to multiply all terms together to get #14sqrt(81)=14*9=126# ... which would be incorrect.

To keep the validity of the complex number system, always deal with the imaginary part first. To do this, rewrite the equation:

#(7sqrt(3)*sqrt(-1))(2sqrt(27)*sqrt(-1))=(sqrt(-1)*sqrt(-1))(7sqrt(3)*2sqrt(27))#

Now you can multiply terms within the parentheses.

#(sqrt(-1)*sqrt(-1))=i*i=i^2=-1#

and

#(7sqrt(3)*2sqrt(27))=14sqrt(81)=14*9=126#

Multiply these together and you get #126*-1=-126#

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