Rationalize the denominator

[CB1101]

Rationalize the denominator:
2/√3 + √2

My work so far:

2/(√3 + √2)
= 2(√3 - √2)

Is 2(√3 - √2) all it's asking for or do I need to do more?

 

[serds]

Is it

(1):

or

(2):

??

 

[CB1101]

Is it

(1):

or

(2):

??

The second one. Sorry for the confusion

 

[Deleted member 4993]

The second one. Sorry for the confusion

To rationalize the denominator, you need to multiply (and divide) by the "conjugate" of the denominator. In this case, you would multiply the expression by

\(\displaystyle \displaystyle \frac{\sqrt{3} \ - \ \sqrt{2}}{\sqrt{3} \ - \ \sqrt{2}} \) - so we have

\(\displaystyle \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \ * \ \frac{\sqrt{3} \ - \ \sqrt{2}}{\sqrt{3} \ - \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2 \ * \ (\sqrt{3} \ - \ \sqrt{2})}{(\sqrt{3} \ + \ \sqrt{2}) \ * \ (\sqrt{3} \ - \ \sqrt{2})} \)

and continue....

 

[Jason76]

To rationalize the denominator, you need to multiply (and divide) by the "conjugate" of the denominator. In this case, you would multiply the expression by

\(\displaystyle \displaystyle \frac{\sqrt{3} \ - \ \sqrt{2}}{\sqrt{3} \ - \ \sqrt{2}} \) - so we have

\(\displaystyle \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \ * \ \frac{\sqrt{3} \ - \ \sqrt{2}}{\sqrt{3} \ - \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2 \ * \ (\sqrt{3} \ - \ \sqrt{2})}{(\sqrt{3} \ + \ \sqrt{2}) \ * \ (\sqrt{3} \ - \ \sqrt{2})} \)

and continue....

\(\displaystyle \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \ * \ \frac{\sqrt{3} \ - \ \sqrt{2}}{\sqrt{3} \ - \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2 \ * \ (\sqrt{3} \ - \ \sqrt{2})}{(\sqrt{3} \ + \ \sqrt{2}) \ * \ (\sqrt{3} \ - \ \sqrt{2})} \)


\(\displaystyle \dfrac{2(\sqrt{3} - \sqrt{2}) }{3 - 2}\)

\(\displaystyle \dfrac{2(\sqrt{3}- \sqrt{2})}{1}\)

\(\displaystyle \dfrac{2(\sqrt{3}) - 2(\sqrt{2})}{1}\)

\(\displaystyle \dfrac{\sqrt{3} \sqrt{2}}{1}\)

\(\displaystyle \sqrt{3} \sqrt{2}\)

 

[Deleted member 4993]

\(\displaystyle \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \ * \ \frac{\sqrt{3} \ - \ \sqrt{2}}{\sqrt{3} \ - \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2 \ * \ (\sqrt{3} \ - \ \sqrt{2})}{(\sqrt{3} \ + \ \sqrt{2}) \ * \ (\sqrt{3} \ - \ \sqrt{2})} \)


\(\displaystyle \dfrac{2(\sqrt{3} - \sqrt{2}) }{3 - 2}\)

\(\displaystyle \dfrac{2(\sqrt{3}- \sqrt{2})}{1}\)

\(\displaystyle \dfrac{2(\sqrt{3}) - 2(\sqrt{2})}{1}\)

\(\displaystyle \dfrac{\sqrt{3} \sqrt{2}}{1}\) .......................... Incorrect - how did get here from above?

\(\displaystyle \sqrt{3} \sqrt{2}\).......................... Incorrect - because line above is incorrect

.

 

[JeffM]

\(\displaystyle \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2}{\sqrt{3} \ + \ \sqrt{2}} \ * \ \frac{\sqrt{3} \ - \ \sqrt{2}}{\sqrt{3} \ - \ \sqrt{2}} \)

\(\displaystyle = \ \displaystyle \frac{2 \ * \ (\sqrt{3} \ - \ \sqrt{2})}{(\sqrt{3} \ + \ \sqrt{2}) \ * \ (\sqrt{3} \ - \ \sqrt{2})} \)


\(\displaystyle \dfrac{2(\sqrt{3} - \sqrt{2}) }{3 - 2}\)

\(\displaystyle \dfrac{2(\sqrt{3}- \sqrt{2})}{1}\)

\(\displaystyle \dfrac{2(\sqrt{3}) - 2(\sqrt{2})}{1}\)

\(\displaystyle \dfrac{\sqrt{3} \sqrt{2}}{1}\) WHAT ARE YOU DOING?

\(\displaystyle \sqrt{3} \sqrt{2}\)

.
\(\displaystyle \sqrt{2}\ and\ \sqrt{3}\) are numbers.

\(\displaystyle 2(12) - 2(7) = 24 - 14 = 10 \ne 84 = 12 * 7.\) Right?

Why in the world would you think that \(\displaystyle 2\sqrt{3} - 2\sqrt{2} = \sqrt{3} * \sqrt{2}\)?

 

[Jason76]

.
\(\displaystyle \sqrt{2}\ and\ \sqrt{3}\) are numbers.

\(\displaystyle 2(12) - 2(7) = 24 - 14 = 10 \ne 84 = 12 * 7.\) Right?

Why in the world would you think that \(\displaystyle 2\sqrt{3} - 2\sqrt{2} = \sqrt{3} * \sqrt{2}\)?

So \(\displaystyle 2\sqrt{3}\) etc.. cannot be seperated?

 

[Deleted member 4993]

So \(\displaystyle 2\sqrt{3}\) etc.. cannot be seperated?

\(\displaystyle 2\sqrt{3} \ = 2 \ * \ \sqrt{3} \ =\ \sqrt{3} \ + \sqrt{3} \ \)

Now what?

What do you exactly mean by separated?

 

[JeffM]

So \(\displaystyle 2\sqrt{3}\) etc.. cannot be seperated?

\(\displaystyle 2\sqrt{3} - 2\sqrt{2} = 2(\sqrt{3} - \sqrt{2})\) if that is what you mean?

It is not generally true that 2a - 2b = ab. What does that have to do with separating anything?

 

[CB1101]

rationalize the denominator

2/(√3 + √2)

____2____ * (√3 - √2)
(√3 + √2).....(√3 - √2)

=
2*(√3 - √2)
(√3 + √2)(√3 - √2)

=
2*(√3 - √2)
3-2+-√6+√6

=
2*(√3 - √2)
Would this be the final answer or is there more to do?
Sorry, I had some trouble with the formatting

 

[stapel]

Looks good to me!

 

[CB1101]

Looks good to me!

ok thanks