Question

Simplify the squareroot of 8/9

Answer 1

Hint

\[\large \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{\sqrt{9}}\]

what do we know?

well we know the square root of 9 is 3 correct?

so right away we can write

\[\large \frac{\sqrt{8}}{3}\]

anything else we can do?

Answer 2

No that's it. Thanks

Answer 3

Not quite :P

Hint...how can we multiply 2 numbers to make 8?

\[\large \frac{\sqrt{2 \times 4}}{3}\]

see anything?

Answer 4

Oh haha, well the sqaureroot of 4 is 2..

Answer 5

Indeed

So the complete simplification is

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

Answer 6

So what if it was negative?

Answer 7

Give me an example?

Answer 8

Squareroot of -9/5

Answer 9

Very important

is it

\[\large \sqrt{\frac{-9}{5}}\]
or
\[\large -\sqrt{\frac{9}{5}}\]

Answer 10

1st one

Answer 11

Question, have you covered imaginary numbers?

Answer 12

Kind of...but it wasn't detailed.

Answer 13

Alright, well as you already know

you cannot take the square root of a negative number...

why? well...what number squared = a negative number? There isnt one...we need to introduce imaginary numbers

Answer 14

So what I like to do...is ignore the fact that we have an '-' there...until the end...you'll see why :)

so lets just solve

\[\large \sqrt{\frac{9}{5}}\]

how can you simplify that? :)

Answer 15

Well the top would be three, but how would you simplify the bottom?

Answer 16

You wouldnt :)

the square root of 5 cannot be simplified

so we just have

\[\large \frac{3}{\sqrt{5}}\]

but remember...we ignored the '-' before

so what do we do now...I'll explain :)

Answer 17

Let me take 1 step back to the beginning

\[\large \frac{\sqrt{-9}}{\sqrt{5}}\]

remember how we broke (in the first problem) the square root of 8? by saying 2 times 4?

well here...think of this as

\[\large \frac{\sqrt{9 \times -1}}{\sqrt{5}}\]

So first....we found that the square root of 9 is 3 right?

\[\large \frac{3\sqrt{-1}}{\sqrt{5}}\]

now, imaginary numbers....you will see that

\[\large \sqrt{-1} = i\]

so our final answer is

\[\large \frac{3i}{\sqrt{5}}\]

Answer 18

the only reason I like to ignore it until the end, is because I know all I have to do is bring along the i at the end

Answer 19

Oh, well I thought in order for radicals to be simplified all the way, there shouldn't be a radical in the denominator...? Last question (:

Answer 20

Yes, I suppose we should :)

sooooo...how would you bring the square root out of the denominator?

Answer 21

I honestly don't know

Answer 22

Recognize the phrase "rationalize the denominator?"

Answer 23

Is it where the denominator only contains rational numbers?

Answer 24

Indeed...

and heres a hint

What is \[\large \sqrt{a} \times \sqrt{a}\]?

Answer 25

That #?

Answer 26

Indeed!!

so here we have

\[\large \frac{3i}{\sqrt{5}}\]

if we multiply the bottom by the square root of 5...we would just have 5 right?

but remember whatever we do to the bottom we do to the top...so
\[\large \frac{3i}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{3i\sqrt{5}}{5}\]

did that make sense?

Answer 27

YES! Thank you so much (:

Answer 28

Anytime :)