Question

Simplifying radical expressions

Answer 1

can you give me the factorization of 128 with one factor being a perfect square?

Answer 2

hang on..

Answer 3

2?

Answer 4

2 * ???

Answer 5

wait im confused

Answer 6

128 = 2 * ???

Answer 7

16?

Answer 8

no... 2 * 16 = 32

Answer 9

64

Answer 10

yes...

Answer 11

\(\large \sqrt{128}=\sqrt{64 \cdot 2}=\sqrt{64} \cdot \sqrt{ 2}=8\sqrt2 \)

now can you do \(\large \sqrt{x^5} \) ????

Answer 12

\[2\sqrt{x}\]

Answer 13

no... how many pairs of x's inside the radical sign? how many left over?

Answer 14

there are 2 pairs and 1 x left over

Answer 15

yes...
so \(\large \sqrt{x^5}=x^2 \sqrt{x} \)

Answer 16

ohh. alright, my bad. i see what i did wrong

Answer 17

i'm thinking that's what you meant to write huh?

Answer 18

lol, yes

Answer 19

ok, now how 'bout \(\large \sqrt{y^2}= \) ????

Answer 20

i dont know what to do there...

Answer 21

how many pairs of y's inside the radical?

Answer 22

1? or would it be 2?

Answer 23

1 pair.... so write one y OUTSIDE the radical

Answer 24

\(\large \sqrt{y^2}=y\sqrt1=y \)

Answer 25

so now let's put all this together to get your answer...

Answer 26

alright.

Answer 27

\(\large \sqrt{128x^5y^2}=\sqrt{128} \cdot \sqrt{x^5} \cdot \sqrt{y^2} \)
\(\large =8\sqrt{2} \cdot x^2\sqrt{x} \cdot y \)
\(\large =8x^2y\sqrt{2} \cdot \sqrt{x} \)
\(\large =8x^2y\sqrt{2x} \)

Answer 28

ok. now just one last one...finally lol

Answer 29

new post pls... :)

Answer 30

willllll do :)