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:

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

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

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

$= 14i + 24i$

$= 34i$

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