Question

@genius12
@eSpeX

Answer 1

rationalize the denominator of \[\sqrt{-36} \over(2-3i)+(3+2i)\]

Answer 2

add them, then multiply by the conjugate

Answer 3

Start simplifying; first note $$\sqrt{-36}=6i$$and then combine our denominator terms:$$2-3i+3+2i=5-i$$... so we now have:$$\frac{6i}{5-i}$$To rationalize, multiply top and bottom by the denominator's *conjugate* -- in this case, \(5+i\):$$\frac{6i}{5-i}\times\frac{5+i}{5+i}=\frac{6i(5+i)}{(5-i)(5+i)}=\frac{6i^2+30i}{25-i^2}=\frac{-6+30i}{26}$$

Answer 4

Observe we have a common factor of \(2\) in both; cancelling it out we end up with$$\frac{-6+30i}{26}=\frac{2(-3+15i)}{2(13)}=\frac{-3+15i}{13}$$

Answer 5

Which of the following represents in radical form? 7x^4/5

Answer 6

4^ sqrt 7x^5 ?

Answer 7

$$7x^{4/5}=7(x^4)^{1/5}=7\sqrt[5]{x^4}$$

Answer 8

Rationalize the denominator of \[3 \over (4+\sqrt5)+(3-2\sqrt5)\]

Answer 9

@oldrin.bataku

Answer 10

@genius12

Answer 11

actually oldrin.bataku already solved it :/

Answer 12

@topstryker follow the steps I outlined above; combine your terms in the denominator, and then multiply by top and bottom by its conjugate!

Answer 13

21+3sqrt5 over 44 ?

Answer 14

@oldrin.bataku