Question

How can you write the expression with a rationalized denominator?

Answer 1

to get a rationalized denominator, you need to multiply both numerator and denominator by a factor which will get rid of the radical sign in the denominator. Do you have any idea what that might be?

Answer 2

No......

Answer 3

Okay. Would you know what to do if that looked like

\[\frac{4}{\sqrt{2}}\]?

Answer 4

no :(

Answer 5

Okay. Just trying to calibrate my description of how to do this.

Do you understand what \(\sqrt{2}\) means? How about \(\sqrt[3]{2}\)?

Answer 6

4 * sqrt{2}
sqrt{2} * sqrt{2}
4*sqrt{2}/2
4/2 = 2
2 sqrt{2}
\(\sqrt[3]{2}\) means cube root of two

Answer 7

Okay. Let's do a simple one first:

\[\frac{4}{\sqrt{2}}\]To rationalize this, we need to multiply the denominator by something that will change that \(\sqrt{2}\) into a rational number (for example, 2). To keep the value the same, we will multiply the numerator by that same something.

What is \(\sqrt{2}*\sqrt{2}=\)?

Answer 8

The squareroots cancel out and we are left with 2

Answer 9

Yes! So if we took our initial fraction and multiplied it by \(\frac{\sqrt{2}}{\sqrt{2}}\) we would not change the value, but we would eliminate the radical from the denominator, rationalizing it.

\[\frac{4}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{4*\sqrt{2}}{\sqrt{2}*\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}\]

As a check, \(\sqrt{2} \approx 1.414\) so we can find the value both ways and compare:
\[\frac{4}{\sqrt{2}} = \frac{4}{1.414} \approx 2.828\]
\[2\sqrt{2} \approx 2*1.414 = 2.828 \checkmark \]
So we rationalized our denominator successfully without changing the value of the fraction. Does that all make sense so far?

Answer 10

Yes :)

Answer 11

Great! So we need to extend that to doing it with a cube root instead of a square root.

What is \(\sqrt[3]{2}*\sqrt[3]{2}*\sqrt[3]{2}=\)

Answer 12

\[\sqrt[3]{2}*\sqrt[3]{2}*\sqrt[3]{2}=2\]
\[\sqrt[4]{2}*\sqrt[4]{2}*\sqrt[4]{2}*\sqrt[4]{2}=2\]

Answer 13

You got it. So if we have a cube root in the denominator, how many times do we have to multiply by the cube root to rationalize it away?

Answer 14

Three times.

Answer 15

Close, but no cigar. That would give us

\[\frac{4}{\sqrt[3]{2}}*\frac{\sqrt[3]{2}}{\sqrt[3]{2}}*\frac{\sqrt[3]{2}}{\sqrt[3]{2}}*\frac{\sqrt[3]{2}}{\sqrt[3]{2}}\]

We want multiply only twice: 1 less than the order of the root. So with a square root, we multiply once. Cube root, twice, 4th root, 3 times, etc.

Answer 16

Oh okay, I get that.

Answer 17

So our original problem was \[\frac{4+\sqrt[3]{3}}{\sqrt[3]{6}}\]We need to multiply by \(\dfrac{\sqrt[3]{6}*\sqrt[3]{6}}{\sqrt[3]{6}*\sqrt[3]{6}}\) to rationalize it.

Answer 18

What do you get if you do that? Denominator will be 6, that part is easy, so I'll do it :-)

Answer 19

So the numerator will be
\(4+\sqrt[3]{3}\times\sqrt[3]{3}\times\sqrt[3]{3}\)
4 + 6 = 10 ?

Answer 20

No, not quite. Remember the distributive property of multiplication?

\[(a+b) * c = a*c + b*c\]

So you need to multiply \[(4+\sqrt[3]{3})*\sqrt[3]{3}*\sqrt[3]{3} = 4*\sqrt[3]{3}*\sqrt[3]{3} + \sqrt[3]{3}*\sqrt[3]{3}*\sqrt[3]{3} =\]

Answer 21

Uh, except those are supposed to be \(\sqrt[3]{6}\) not \(\sqrt[3]{3}\)!

Answer 22

\[(4+\sqrt[3]{3})*\sqrt[3]{6}*\sqrt[3]{6} = 4*\sqrt[3]{6}*\sqrt[3]{6} + \sqrt[3]{6}*\sqrt[3]{6}*\sqrt[3]{6} =\]

Answer 23

\[4*\sqrt[3]{6}*\sqrt[3]{6} + \sqrt[3]{6}*\sqrt[3]{6}*\sqrt[3]{6} =\]
\[4*\sqrt[3]{36} + 6 \]
\[10*\sqrt[3]{36} \]

Answer 24

Argh, we still don't have it right :-)

\[(4+\sqrt[3]{3})*\sqrt[3]{6}*\sqrt[3]{6} = 4*\sqrt[3]{6}*\sqrt[3]{6} + \sqrt[3]{3}*\sqrt[3]{6}*\sqrt[3]{6}*\sqrt[3]{6} =\]

Answer 25

And you can't add like that, by the way — you'll have two terms in the answer

Answer 26

\[(4+\sqrt[3]{3})*\sqrt[3]{6}*\sqrt[3]{6} = 4*\sqrt[3]{6}*\sqrt[3]{6} + \sqrt[3]{3}*\sqrt[3]{6}*\sqrt[3]{6} = 4\sqrt[3]{36}+\sqrt[3]{108} \]\[= 4\sqrt[3]{36}+\sqrt[3]{27*4}=4\sqrt[3]{36}+3\sqrt[3]{4}\]

That's our numerator, don't forget the denominator!