Question

Simplify by rationalizing each denominator.. Square root of 11 over 5 times the square root of 132

Answer 1

\[\frac{\sqrt{11}}{5\sqrt{132}}\]?

Answer 2

Yes

Answer 3

ok
since \(132=2^2\times 3\times 11\) if you multiply top and bottom by \(\sqrt{3\times 11}\) aka \(\sqrt{33}\) you will have a perfect square in the denominator

Answer 4

\[\frac{\sqrt{11}}{5\sqrt{132}}\times\frac{\sqrt{33}}{{\sqrt{33}}}\]is a start

Answer 5

then the denominator will be
\[5\sqrt{2^2\times 3^2\times 11^2}=5\times 2\times 3\times 11=330\]

Answer 6

and the numerator will be \(11\sqrt3\) then you can cancel an 11 top and bottom

Answer 7

Im confused.

Answer 8

ok lets go slow

Answer 9

if you had \[\frac{2}{\sqrt3}\] and you wanted to rationalize the denominator you would multiply top and bottom by \(\sqrt3\) and get \[\frac{2}{\sqrt3}\times \frac{\sqrt3}{\sqrt3}=\frac{2\sqrt3}{3}\]

Answer 10

is that confusing or is that clear?

Answer 11

Well i understand how to do that one, but its the 5 thats throwing me off .-.

Answer 12

ignore the 5, just leave it there

Answer 13

suppose you had
\[\frac{7}{\sqrt{12}}\] and you wanted to rationalize the denominator
what you would multiply top and bottom by ?

Answer 14

The square root of 12

Answer 15

you could but you do not need to, that is the point here
\[12=4\times 3\] and \(\sqrt{4}=2\)
so you could just multiply by \(\sqrt3\) top and bottom and get
\[\frac{7}{\sqrt{12}}\times \frac{\sqrt3}{\sqrt3}=\frac{7\sqrt3}{\sqrt{36}}=\frac{7\sqrt3}{6}\]

Answer 16

So you multiply the top & bottom by the multiple of the bottom that doesnt have a square root?

Answer 17

exactly !

Answer 18

i mean you could do it the other way as well, but then you would have a bunch to cancel at the end

Answer 19

that is why the very first thing i wrote was \[132=2^2\times 3\times 11\] so show that you only have to multiply by \(\sqrt{11\times 3}\)

Answer 20

But in the original question the square root of 11 over 5 times the share root of 132, you multiply the top & bottom but the square root of 33 which gives you 363 on top & 5 times the square root of 4356, thats a big number is that wrong what next? .-.

Answer 21

if you leave it in factored form it is much easier

Answer 22

\[\frac{\sqrt{11}}{5\sqrt{2^2\times3\times 11}}\times\frac{\sqrt{33}}{{\sqrt{3\times 11}}}\]

Answer 23

then you see right away that the denominator is
\[5\times 2\times 3\times 11\]

Answer 24

and the numerator is
\[\sqrt{11\times 11\times 3}=11\sqrt3\]

Answer 25

Im never going to get this! 😥

Answer 26

it is not that bad really
just don't multiply out
the whole point is to get a perfect square in the denominator
i have no idea what
\[2^2\times3^2\times 11^2\]is but i am certain that its square root is \(2\times 3\times 11\)

Answer 27

But why 2 if its 2^2

Answer 28

what i meant was
\[\sqrt{2^2\times 3^2\times 11^2}\]is
\[2\times 3\times 11\]
clear or no?

Answer 29

Wait! I think i have it!

Answer 30

If the denominator is 5 times the square root of 2^2 x 3 x11 and your multiplying by 3 x 11 it turns into 5 x 2^2 x 3^3 x 11^11 then the square roots of that are 5x2x3x11 ?

Answer 31

yes yes yes
and yes

Answer 32

But then whats the answer i dont know where to go from that

Answer 33

ok we write it
second to last step is
\[\frac{\sqrt{11}\sqrt{3\times11}}{5\times 2\times 3\times 11}\]

Answer 34

then the numerator is \(11\sqrt3\) so we get
\[\frac{11\sqrt3}{5\times 2\times 3\times 11}=\frac{\sqrt3}{5\times 2\times 3}\]

Answer 35

notice the 11 canceled
if we had been paying attention we could have cancelled it first , but no matter