Question

2sqaure root6 /2squre root2 - 2sqaure root2/2sqaure root 2 =?

Answer 1

let me clarify the question.... is it like this:

\(\large \frac{2\sqrt6}{2\sqrt2}-\frac{2\sqrt2}{2\sqrt2} \)

????

Answer 2

Just combine them..

Answer 3

As the denominator is same, just combine them..

And subtract the numerator..

Answer 4

yh i did but with the denominator i always get confused with removing the root signs

Answer 5

Don't worry, talented @dpaInc is here to help you..

Answer 6

yes... denominators are the same, so just subtract the numerator...
factor out a 2 from the numerator and cancel with the 2 in the denominator...

but not as talented as @waterineyes , !!!

Answer 7

awww... what happened? i was waiting for a response....

Answer 8

\[2(\sqrt{6} - \sqrt{2}) / 2\sqrt{2} - \sqrt{2})\]

Answer 9

like that ?

Answer 10

waterineyes is talented??

Really?

Breaking News for today..

Answer 11

the denominator is incorrect....
here...
\(\large \frac{2\sqrt6}{2\sqrt2}-\frac{2\sqrt2}{2\sqrt2}=\frac{2\sqrt6-2\sqrt2}{2\sqrt2}=\frac{2(\sqrt6-\sqrt2)}{2\sqrt2}=\frac{\cancel2^1(\sqrt6-\sqrt2)}{\cancel2^1\sqrt2}=\frac{\sqrt6-\sqrt2}{\sqrt2} \)

now we must rationalize the denominator...

Answer 12

and yes... talented.... it's old news.... :)

Answer 13

how do we know when to rationalize the denominator ?

Answer 14

simplest radical form means there are no radical(s) in the denominator of a rational expression.

Answer 15

i don't get it

Answer 16

See if you have radical in the denominator, then to make the denominator you will rationalize it??

Getting??

Answer 17

yeah okay :)

Answer 18

For example:

\[\frac{1}{\sqrt{2}}\]

Answer 19

Here you can rationalize it...

Answer 20

so then do we get sqaure root 12- sqaure root 4 over 4

Answer 21

hmmm .... now that i think about it, we went about this the long way when we could have just done it from the beginning....

Answer 22

Can't we take sqrt{2} common from numerator? @dpaInc

Answer 23

when you multiply the denominator square root 2 by square root 2 do u just get 2 ? or 2x2

Answer 24

\[\frac{2\sqrt6}{2\sqrt2}-\frac{2\sqrt2}{2\sqrt2}=\frac{2\sqrt6-2\sqrt2}{2\sqrt2}=\frac{2(\sqrt6-\sqrt2)}{2\sqrt2}=\frac{\cancel2^1(\sqrt6-\sqrt2)}{\cancel2^1\sqrt2}=\frac{\sqrt6-\sqrt2}{\sqrt2}\]

Answer 25

\[\frac{\sqrt6-\sqrt2}{\sqrt2} = \frac{\sqrt{2}(\sqrt{3} - 1)}{\sqrt{2}}\]

Answer 26

like this...
\(\large \frac{2\sqrt6}{2\sqrt2}-\frac{2\sqrt2}{2\sqrt2}=\frac{\cancel2\sqrt6}{\cancel2\sqrt2}-1=\frac{\cancel{\sqrt2} \sqrt3}{\cancel{\sqrt2}}-1=\sqrt3 - 1 \)

Answer 27

\[\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \cdot \sqrt{3}\]

Answer 28

haha.... many ways to do it...
what I tell u??? the talented @waterineyes .... :)

Answer 29

im really confused now :(

Answer 30

can you please show me one step in order

Answer 31

ok... i think the shortest and fastest way is that last way i showed....
\(\large \frac{2\sqrt6}{2\sqrt2}-\frac{2\sqrt2}{2\sqrt2}=\frac{\cancel2\sqrt6}{\cancel2\sqrt2}-1=\frac{\cancel{\sqrt2} \sqrt3}{\cancel{\sqrt2}}-1=\sqrt3 - 1 \)

do you undestand the flow?

Answer 32

how did u get -1 ?

Answer 33

ohh yhhh i get it

Answer 34

\(\large \frac{2\sqrt2}{2\sqrt2}=1 \)

dooh... you already rplied...

Answer 35

the answer has square root 6 - square root 4 over 4 ?

Answer 36

this is the answer: \(\large \frac{\sqrt6 - \sqrt4}{4} \) ????

Answer 37

yeah its in the text book ?

Answer 38

you sure? i don't think thats right...

Answer 39

lemme double check....

Answer 40

thats what it says

Answer 41

Do you have scanner??
Can you here upload the copy of question ??

Answer 42

well that's not what mr. wolfram says....
http://www.wolframalpha.com/input/?i=%282+sqrt%286%29%29%2F%282+sqrt%282%29%29-%282+sqrt%282%29%29%2F%282+sqrt%282%29%29

Answer 43

If we rationalize it then also we get:

\[\frac{2 \sqrt{3} - 2}{2}\]

Answer 44

its questions 3b

Answer 45

that's a trig expression... not the radical thingy you gave...

Answer 46

yeah if u solve it then u sure get what i gave u

Answer 47

should *

Answer 48

not quite... the first product is sqrt3/2 * sqrt2/2 = sqrt6/4

Answer 49

i got square root 3 over2 x 1pver square root2 - 1/2 x 1/square root 2

Answer 50

ok.. everything is messed can we start with the original question from the book?

Answer 51

okay :)

Answer 52

ok...
here's question 3b:
\(\large sin(\pi/3)cos(\pi/4)-cos(\pi/3)sin(\pi/4) \)
= \(\large (\frac{\sqrt3}{2})(\frac{\sqrt2}{2})-(\frac{1}{2})(\frac{\sqrt2}{2}) \)

is this ok so far?

Answer 53

yeah :)

Answer 54

wait sorry

Answer 55

for sin pi/4 i have 1/sqaure root 2

Answer 56

\(\large \frac{1}{\sqrt2}=\frac{\sqrt2}{2} \) they're the same value only the right side is in simplest radical form

Answer 57

kk

Answer 58

\(\large sin(\pi/3)cos(\pi/4)-cos(\pi/3)sin(\pi/4) \)
= \(\large (\frac{\sqrt3}{2})(\frac{\sqrt2}{2})-(\frac{1}{2})(\frac{\sqrt2}{2}) \)
= \(\large \frac{\sqrt6}{4}-\frac{\sqrt2}{4} \)
= \(\large \frac{\sqrt6-\sqrt2}{4}\)

Answer 59

thank uu

Answer 60

and that answer matches what wolfram alpha has...
but it still does not match your answerbook...
i'm confident that this answer is correct...
yw.... :)

Answer 61

no thats the correct answer

Answer 62

ok...:)