Question

Rationalize the denominator. Assume all expressions under radicals represent positive numbers.

3sqrt5y^4 over 3sqrt10x^4

Answer 1

\[3\sqrt 5y^4 \over 3\sqrt10x^4\]

Answer 2

There we go.

That's what I was looking for. :)

What does it mean to rationalize the denominator?

Answer 3

no clue

Answer 4

I would assume to find the common den.

Answer 5

if you want \[\sqrt{10x^4}\] you can write sqrt{10x^4} in equation editor just so you know

Answer 6

Well, the square root of a non-perfect square is known as an irrational number, so right now the denominator is an irrational number. We want to "rationalize" it by getting rid of that square root.

Answer 7

ok.

Answer 8

so we just have 10x^4?

Answer 9

And is it just the square root of 10 or the expression 10x^4

Answer 10

k

Answer 11

is it \[\sqrt{10}x^4 or \sqrt{10x^4}\]

Answer 12

Thanks myin.

Answer 13

the second one. I messed up

Answer 14

Okay, so what can I multiply that by to get rid of the square root? Any ideas or thoughts?

Answer 15

and i assume its the second way for the numerator well except with different numbers

Answer 16

2?

Answer 17

anyways we want to rationalize the denominator?
so are there any perfect squares under the square root?

Answer 18

I'll give you a hint.

The square root of x is the same as x^(1/2)

If I take x^(1/2) * x^(1/2) what do I get?

Answer 19

Remember what we talked about with exponents on our last problem! :)

Answer 20

0

Answer 21

We are multiplying this time; not dividing.

Answer 22

1

Answer 23

Yes, 1.

So if I took \[\sqrt{x} \times \sqrt{?}\]

Answer 24

Oops. I meant to write the square root of x both times.

Answer 25

x^2?

Answer 26

If I multiply the square root of x by the square root of x I got x, didn't I ?

Answer 27

so it's just x?

Answer 28

Remember x^(1/2) * x^(1/2) = x^(1/2 + 1/2) = x^1 = x

Answer 29

Yes, it's just x. So if I have the square root of 10x^4, what can I multiply that by to get rid of the square root?

Answer 30

-4 idk i'm getting confused again lol

Answer 31

4

Answer 32

Ok.

\[\sqrt{10x^4} = (10x^4)^{1/2}\]

Answer 33

So how could I make sure the exponent equals 1?

Answer 34

no idea = /
my luck, it's probably really simple to figure out

Answer 35

Remember x^(1/2) * x^(1/2) = x

So (10x^4)^(1/2) * (????)^(1/2) = 10x^4

Answer 36

1

Answer 37

Basically, if you have the square root of a value and you multiply it by the square root of that same value, you ALWAYS get the value of what is inside the radical.

Answer 38

\[\sqrt{10x^4} \times \sqrt{10x^4} = 10x^4\]

Answer 39

It works for anything.

Answer 40

ooooh. ok.

Answer 41

It's because you are adding the exponents which are 1/2 and 1/2 + 1/2 = 1

Answer 42

So back to the original question, you would multiply both the numerator and the denominator by sqrt(10x^4)

Answer 43

The first 2 steps are actually easier than this.

Can we cancel out the 3s?

Can we do anything with the square root of the variables?

Answer 44

cancel them out as well?

Answer 45

We can cancel out the 3s.

However, if we look at x^4 that's the same as x * x * x * x. How many groups of 2 Xs are there?

Answer 46

2

Answer 47

\[\sqrt{x^4} = x^2\]

Answer 48

ok

Answer 49

Exactly! So we can rewrite this as the following:

\[y^2\sqrt{5} / x^2\sqrt{10}\]

Answer 50

So now all we need to do is multiply the numerator and denominator by what to get rid of that sqrt(10) ?

Answer 51

x^2sqrt10?

Answer 52

Close. Just sqrt(10)

Answer 53

ok

Answer 54

The x^2 is fine on its own

Answer 55

So if you multiply the top and bottom by that, what are you left with?

Answer 56

5y and x^4?

Answer 57

y^2 * sqrt(5) * sqrt(10) / x^2 * sqrt(10) * sqrt(10)

y^2 * sqrt(50) / x^2 * 10

Answer 58

oh

Answer 59

\[y^2\sqrt{5}\sqrt{10} = y^2\sqrt{50}\]

Answer 60

\[x^2\sqrt{10}\sqrt{10} = 10x^2\]

Answer 61

Can you simplify sqrt(50) any further?

Answer 62

sqrt10?

Answer 63

50 = 2 x 5 x 5

Answer 64

ok

Answer 65

So I can pull out a 5 and am left with

\[5\sqrt{2}\]

Answer 66

2*5

Answer 67

The 5 then will simplify with the 10 in the denominator leaving a 2 in the denominator.

Answer 68

oooh

Answer 69

So..

We are left with

\[y^2\sqrt{2} / 2x^2\]

Answer 70

No more irrational stuff in the denominator!

Answer 71

awesome!

Answer 72

Yeah, if you're into that sort of thing.