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

Answer 1

Theodore Amidon

#=+-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#

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Answer 2

Chloe Alexander

#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#

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Answer 3

Madelyn Anderson

To simplify (2\sqrt{-49} + 3\sqrt{-64}), first, simplify the square roots:

[ \sqrt{-49} = \sqrt{(-1)(49)} = \sqrt{-1} \times \sqrt{49} = 7i ]

[ \sqrt{-64} = \sqrt{(-1)(64)} = \sqrt{-1} \times \sqrt{64} = 8i ]

Then substitute the simplified values back into the expression:

[ 2\sqrt{-49} + 3\sqrt{-64} = 2(7i) + 3(8i) = 14i + 24i = 38i ]

So, (2\sqrt{-49} + 3\sqrt{-64}) simplifies to (38i).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.