How would you solve √-3 * √-27 and 12i / 4i ?

(√-3) * (√-27) = (√[3*(-1)] ) * (√[27*(-1)] ) = (√3)*(√-1) * (√27)*(√-1)

.... (√-1) * (√-1) = -1 .... definition

(√3)*(√27)*(-1) = -√(3*27) = -√81

They may be looking for the -9 or choice (a.) .... but

technically √81 has two solutions = +9 and -9

(9) * (9) = +81

(-9) * (-9) = +81

and so ... -√81 has the same two solutions = (-1)*(9) and (-1)*(-9) = -9 and +9

So the true answer is (a.) and (b.)

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(12i) / (4i) = 3 .... or choice (c.) .... Its just like (12x) / (4x) = 3

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Useful information:

(√-1) = i

i * i = i² = -1

i^4 = 1

first q -

√-3*√-27 = i^2√81 = -9
( because i^2 = -1)




second q -

12i / 4i = 12 / 4 = 3

Qu 1
√ ( 3 i ² ) √ ( 27 i ² )
i √3 i 3 √3
i ² 9
- 9
OPTION a

Qu 2

12 i
-------
4 i

3

OPTION c

1)
√(-3) * √(-27)
= √(-3 * -27)
= √81
= 9
(answer b)

2)
12i/4i
= 12/4
= 3
(answer c)

√-3 * √-27 =

√-3 = i√3 & √-27 = i√27, so:

i√3 x i√27 = i^2√27(3)
= -1(√81)
= -9

final answer ---> a. -9

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12i / 4i = 3

final answer ---> c. 3