Question

Simplifying radical expressions

Answer 1

let's do it this way.... factor 63....

Answer 2

ok give me a sec

Answer 3

can i do it by 3? or 9?

Answer 4

does not matter.... how 'bout 63 = 9 * 7 ????

Answer 5

that ok???

Answer 6

yes thats thats fine.

Answer 7

ok... so here's the problem:

Answer 8

\(\large 8\sqrt{63x^5}=8\sqrt{63} \sqrt{x^5}=8\sqrt{9\cdot 7} \sqrt{x^5}=8\sqrt{9}\cdot \sqrt{ 7} \sqrt{x^5} \)
= \(\large 8 \cdot 3 \cdot \sqrt7\cdot \sqrt{x\cdot x\cdot x\cdot x\cdot x} \)

Answer 9

are we good so far???? lemme know if you don't understand anything.....

Answer 10

yes im understanding this i think.

Answer 11

"i think.." ??? you have doubts??? where?

Answer 12

i mean I am understanding :) sorry haha

Answer 13

ok good.. so this is what we have so far (from that last line):
\(\large 8\cdot 3 \cdot \sqrt7 \cdot \sqrt{x \cdot x \cdot x \cdot x \cdot x} \)
= \(\large 24 \cdot \sqrt7 \cdot \sqrt{x \cdot x \cdot x \cdot x \cdot x} \)

let's just work with that last radical... how many pairs of x's are there in the radicand?

Answer 14

2 pairs

Answer 15

right... 2 pairs with one left over.... correct????

Answer 16

yes. that is correct

Answer 17

so i can write two x's OUTSIDE of the radical and leave the remaining one inside:
\(\large \sqrt {x \cdot x \cdot x \cdot x \cdot x }=x \cdot x\sqrt {\cancel{x \cdot x} \cdot \cancel{ x \cdot x} \cdot x }=x^2\sqrt{x} \)

we good still?

Answer 18

yes, i was just making sure i understood everything so far sorry it took a min

Answer 19

np....
so now putting this into the original problem:
\(\large 24\sqrt7\cdot x^2 \sqrt{x}=24x^2\sqrt{7} \sqrt{x}=24x^2\sqrt{7 x} \)

Answer 20

ok now that is the answer?

Answer 21

yep.... it is...

Answer 22

I didnt want to get the steps confused with the answer, thats why i was asking

Answer 23

ok i have two more..

Answer 24

ok... new post pls... :)

Answer 25

great. Will do :)