How can I rationalize the numerator in this fraction with square roots?

  • Thread starter JettyZ
  • Start date
In summary: So, what I'm saying is, you should really check your work harder before you post it.But it's not my work, it's your work. I'm confused. What exactly are you talking about?I see, you are right. I was thinking of the wrong thing, sorry. So we end up with the same result.In summary, the problem given is to rationalize the numerator of the fraction (1+sqrt(2)-sqrt(3))/(sqrt(2)), and the solution is to first separate the fraction into three fractions, rationalize the numerator of each fraction, and then combine them back together. By using the conjugate of the numerator, we can easily rationalize the numerator of each fraction. The
  • #1
JettyZ
8
0
Alright, so I've go this question where I'm to rationalize the numerator. [Grade 10 Level]

Homework Statement


(1+sqrt(2)-sqrt(3))/(sqrt(2))
http://www.myalgebra.com/math_image.aspx?p=SMB02FSMB031+SMB02RSMB032SMB02rSMB03-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03?p=74?p=42 [Broken]

Homework Equations


The Attempt at a Solution


I'm not going to type all of my entire failed attempts. Though, I did try seperating this into two fractions, but I don't think that'd be useful. I also multiplied the numerator and denominator by the numerator's conjugate, I think. Although I'm not sure:
(1+sqrt(2)-sqrt(3))/(sqrt(2))*((sqrt(2))+(sqrt(3)))/((sqrt(2))+(sqrt(3)))
http://www.myalgebra.com/math_image.aspx?p=SMB02FSMB031+SMB02RSMB032SMB02rSMB03-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03*SMB02FSMB03(SMB02RSMB032SMB02rSMB03)+(SMB02RSMB033SMB02rSMB03)SMB10(SMB02RSMB032SMB02rSMB03)+(SMB02RSMB033SMB02rSMB03)SMB02fSMB03?p=164?p=42 [Broken]
I think the positive one in the equation is what's causing me to lose.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
JettyZ said:
Alright, so I've go this question where I'm to rationalize the numerator. [Grade 10 Level]

Homework Statement


(1+sqrt(2)-sqrt(3))/(sqrt(2))
http://www.myalgebra.com/math_image.aspx?p=SMB02FSMB031+SMB02RSMB032SMB02rSMB03-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03?p=74?p=42 [Broken]


Homework Equations





The Attempt at a Solution


I'm not going to type all of my entire failed attempts. Though, I did try seperating this into two fractions, but I don't think that'd be useful. I also multiplied the numerator and denominator by the numerator's conjugate, I think. Although I'm not sure:
(1+sqrt(2)-sqrt(3))/(sqrt(2))*((sqrt(2))+(sqrt(3)))/((sqrt(2))+(sqrt(3)))
http://www.myalgebra.com/math_image.aspx?p=SMB02FSMB031+SMB02RSMB032SMB02rSMB03-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03*SMB02FSMB03(SMB02RSMB032SMB02rSMB03)+(SMB02RSMB033SMB02rSMB03)SMB10(SMB02RSMB032SMB02rSMB03)+(SMB02RSMB033SMB02rSMB03)SMB02fSMB03?p=164?p=42 [Broken]



I think the positive one in the equation is what's causing me to lose.

Are you sure they don't want you to rationalize the denominator instead? Why would you want to rationalize the numerator of a fraction?
 
Last edited by a moderator:
  • #3
The teacher just told us to... It's homework.
 
  • #4
JettyZ said:
The teacher just told us to... It's homework.

Very weird, but fair enough. I think you had the right idea to separate it into multiple fractions. I'd split it up into all 3 fractions, rationalize the numerator of each, and look for any re-combinations (if there are any) that do not re-introduce radicals in the numerator(s). Not sure that's what they want, but it's all that I see at the moment.
 
  • #5
Oh, so I separated it into three fractions and I think I just won.

http://www.myalgebra.com/math_image.aspx?p=2+SMB02FSMB031-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03?p=66?p=42 [Broken]

Thanks :)
 
Last edited by a moderator:
  • #6
JettyZ said:
Oh, so I separated it into three fractions and I think I just won.

http://www.myalgebra.com/math_image.aspx?p=2+SMB02FSMB031-SMB02RSMB033SMB02rSMB03SMB10SMB02RSMB032SMB02rSMB03SMB02fSMB03?p=66?p=42 [Broken]

Thanks :)

But your answer still has a radical in a numerator... (and I'm not sure your answer is equal to the original problem either... have you tried putting both forumulas into your calculator to check that they are equal?)
 
Last edited by a moderator:
  • #7
berkeman said:
But your answer still has a radical in a numerator... (and I'm not sure your answer is equal to the original problem either... have you tried putting both forumulas into your calculator to check that they are equal?)

Oh... You're right. I guess that's a problem then.
 
  • #8
Hint:

Start with noting:
[tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{(1+(\sqrt{2}-\sqrt{3}))(1-(\sqrt{2}-\sqrt{3}))}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))}[/tex]

The numerator you'll end up with will be easy to rationalize
 
  • #9
arildno said:
Hint:

Start with noting:
[tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{(1+(\sqrt{2}-\sqrt{3}))(1-(\sqrt{2}-\sqrt{3}))}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))}[/tex]

The numerator you'll end up with will be easy to rationalize

Thanks, I noticed that your method would require more steps to complete than:
[tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{((1+\sqrt{2})-\sqrt{3})((1+\sqrt{2})+\sqrt{3})}{\sqrt{2}((1-\sqrt{2})-\sqrt{3})}[/tex]

Anyways so I've solved it, thanks for the help. :)
 
  • #10
As a final note, we may simplify this very much:
[tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{(1+(\sqrt{2}-\sqrt{3}))(1-(\sqrt{2}-\sqrt{3}))}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))}=\frac{2(\sqrt{6}-2)}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))}=\frac{4}{\sqrt{2}(1-(\sqrt{2}-\sqrt{3}))(\sqrt{6}+2)}[/tex]

By multiplying out the denumerator, you'll end up with:
[tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{2}{1+\sqrt{2}+\sqrt{3}}[/tex]
 
  • #11
JettyZ said:
Thanks, I noticed that your method would require more steps to complete than:
[tex]\frac{1+\sqrt{2}-\sqrt{3}}{\sqrt{2}}=\frac{((1+\sqrt{2})-\sqrt{3})((1+\sqrt{2})+\sqrt{3})}{\sqrt{2}((1-\sqrt{2})-\sqrt{3})}[/tex]

Anyways so I've solved it, thanks for the help. :)
This won't work!

Unless the -signs in the denumerator really should be +signs, in which case it works nicely.
 

1. What is "rationalizing the numerator"?

"Rationalizing the numerator" is a mathematical process that involves simplifying a fraction by removing any radicals (square roots or cube roots) from the numerator and placing them in the denominator.

2. Why is it important to rationalize the numerator?

Rationalizing the numerator can make it easier to perform mathematical operations on the fraction, as well as make it easier to compare fractions or write them in simplest form.

3. How do you rationalize the numerator?

To rationalize the numerator, you need to multiply both the numerator and denominator of the fraction by a suitable term that will eliminate the radical in the numerator. This can be done by multiplying by the conjugate of the radical in the numerator.

4. Can you give an example of rationalizing the numerator?

Sure, for example, if we have the fraction 3/√2, we can rationalize the numerator by multiplying both the numerator and denominator by √2. This gives us (3*√2)/(√2*√2) = (3√2)/2. Now, the radical is in the denominator and the fraction is rationalized.

5. Is rationalizing the numerator always necessary?

No, rationalizing the numerator is not always necessary, but it can be helpful in certain situations, such as when simplifying complex fractions or when comparing fractions with different radicals in the numerator.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
779
  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
12K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
17K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
908
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top