Math Homework Help Needed!

thegreatjeff

Senior member
Jun 14, 2000
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OK, I have some homework due in a little while (actually on Wednesday lol), and I just *cannot* work it out, I need some help...
ok here it is:

Directions: Rationalize the Denominator and Simplify:
1
-----------
1+sq. root of 2+sq.root of 3
And the math teacher said it had something to do with conjugates, but whenever I try that, it won't work...so if you could give me something to go off of, that will help me do it, I'd greatly appreciate it! =)
Thanx
EDIT: the square root symbols weren't showing up correctly, i fixed em
 

rgwalt

Diamond Member
Apr 22, 2000
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I would help, but I turned my math brain off today after a big presentation I gave on bifurcation analysis of predator/prey models. I will do no math for the next couple of weeks.

But, heres a bump for you...
 

thegreatjeff

Senior member
Jun 14, 2000
914
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0
Originally posted by: rgwalt
I would help, but I turned my math brain off today after a big presentation I gave on bifurcation analysis of predator/prey models. I will do no math for the next couple of weeks.

But, heres a bump for you...
Aggh...no! help me! Hahaha...thanx for the bump anyway
 

rob3rt

Member
Jun 7, 2001
114
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Heres a hint:

Think of the denominator as (1+sqrt(2))+sqrt(3) and multiply it by its conjugate. This will give you a fraction with only one irrational number in the denominator. Repeat the process again (with conjugates or other methods), and you will have your answer. :)

The answer might not be as pretty as the original, but at least it won't have a denominator with an irrational.
 

MacBaine

Banned
Aug 23, 2001
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Originally posted by: thegreatjeff
OK, I have some homework due in a little while (actually on Wednesday lol), and I just *cannot* work it out, I need some help...
ok here it is:

Directions: Rationalize the Denominator and Simplify:
1
-----------
1+sq. root of 2+sq.root of 3
And the math teacher said it had something to do with conjugates, but whenever I try that, it won't work...so if you could give me something to go off of, that will help me do it, I'd greatly appreciate it! =)
Thanx
EDIT: the square root symbols weren't showing up correctly, i fixed em
Is it written as you wrote it on the bottom exactly, or are there parenthases... it is kinda confusing.

Is it:

1+(root2)+(root3)
or
1+(root(2+root3))
or what?

Clarify and I can help you


 

thegreatjeff

Senior member
Jun 14, 2000
914
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0
OK Everyone, thanx for your help...first of all it's supposed to be written like this:
1
---------
1 + (sq. root of 2) + (sq. root of 3)

I multiplied by the conjugate, and came up with:

-1 - (sq. root of 2) - (sq. root of 3)
--------------------------------------------
1 - 2 - (sq. root of 6) - (sq. root of 6) - 3

And that's correct, becuz I figured it out on my calculator, now I just need to get rid of the *square root of 6* terms...

Now I'm stuck...again haha...I tried multiplying [1 - 2 - (sq. root of 6) - (sq. root of 6) - 3] by it's conjugate, [1 + 2 + (sq. root of 6) + (sq. root of 6) + 3] (as well as the numerator, but came up with the wrong answer, I still had some irrational numbers in the denominator..?

On a related note, what is the conjugate of (-1 + 2)? Is it (-1 - 2) or (+1 - 2)? EDIT: OK I figured it out myself, it's (-1 - 2)...I still need help with the main problem though...:)
 

Alphathree33

Platinum Member
Dec 1, 2000
2,419
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I don't know what ya'll are smoking but this is fairly easy. It was designed to work out nicely.

Look,

1 / (1 + root2 + root3)

= 1 / [(1 + root2) + root3]

= [1 + root2 - root3] / [(1 + root2)^2 - 3] (multiplied by the conjugate of the bottom, taken as two terms not three)

= [1 + root2 - root3] / [1 + 2root2 + 2 - 3]

= [1 + root2 - root3] / [2root2]

= [root2 + 2 - root6] / 4

Done.
 

thegreatjeff

Senior member
Jun 14, 2000
914
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Originally posted by: Alphathree33
I don't know what ya'll are smoking but this is fairly easy. It was designed to work out nicely.

Look,

1 / (1 + root2 + root3)

= 1 / [(1 + root2) + root3]

= [1 + root2 - root3] / [(1 + root2)^2 - 3] (multiplied by the conjugate of the bottom, taken as two terms not three)

= [1 + root2 - root3] / [1 + 2root2 + 2 - 3]

= [1 + root2 - root3] / [2root2]

= [root2 + 2 - root6] / 4

Done.
Sorry, that doesn't work...the original problem, when divided, was roughly .24118, and that answer equals roughly 1.4659. The problem comes in after you mulptiplied by the conjugate, because on line four of your answer, it should look like this once you square the (1+root2):
= [1 + root2 - root3] / [1 + (sq. root of 2) + (sq. root of 2) + (sq. root of 2) + (sq. root of 2)]
or...
= [1 + (sq. root of 2) + (sq. root of 3)] / [1 + (sq. root of 2)(4)]
Correct me if I'm wrong, I think I'm not getting something in your answer...
 

Rahminator

Senior member
Oct 11, 2001
726
0
0
Originally posted by: thegreatjeff
Originally posted by: Alphathree33
I don't know what ya'll are smoking but this is fairly easy. It was designed to work out nicely.

Look,

1 / (1 + root2 + root3)

= 1 / [(1 + root2) + root3]

= [1 + root2 - root3] / [(1 + root2)^2 - 3] (multiplied by the conjugate of the bottom, taken as two terms not three)

= [1 + root2 - root3] / [1 + 2root2 + 2 - 3]

= [1 + root2 - root3] / [2root2]

= [root2 + 2 - root6] / 4

Done.
Sorry, that doesn't work...the original problem, when divided, was roughly .24118, and that answer equals roughly 1.4659. The problem comes in after you mulptiplied by the conjugate, because on line four of your answer, it should look like this once you square the (1+root2):
= [1 + root2 - root3] / [1 + (sq. root of 2) + (sq. root of 2) + (sq. root of 2) + (sq. root of 2)]
or...
= [1 + (sq. root of 2) + (sq. root of 3)] / [1 + (sq. root of 2)(4)]
Correct me if I'm wrong, I think I'm not getting something in your answer...
He's right. You get 1.4659 because you're forgetting parenthesis somewhere. Enter it like this: (sqrt(2) + 2 - sqrt(6)) / 4
 

Alphathree33

Platinum Member
Dec 1, 2000
2,419
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Originally posted by: Rahminator
Originally posted by: thegreatjeff
Originally posted by: Alphathree33 I don't know what ya'll are smoking but this is fairly easy. It was designed to work out nicely. Look, 1 / (1 + root2 + root3) = 1 / [(1 + root2) + root3] = [1 + root2 - root3] / [(1 + root2)^2 - 3] (multiplied by the conjugate of the bottom, taken as two terms not three) = [1 + root2 - root3] / [1 + 2root2 + 2 - 3] = [1 + root2 - root3] / [2root2] = [root2 + 2 - root6] / 4 Done.
Sorry, that doesn't work...the original problem, when divided, was roughly .24118, and that answer equals roughly 1.4659. The problem comes in after you mulptiplied by the conjugate, because on line four of your answer, it should look like this once you square the (1+root2): = [1 + root2 - root3] / [1 + (sq. root of 2) + (sq. root of 2) + (sq. root of 2) + (sq. root of 2)] or... = [1 + (sq. root of 2) + (sq. root of 3)] / [1 + (sq. root of 2)(4)] Correct me if I'm wrong, I think I'm not getting something in your answer...
He's right. You get 1.4659 because you're forgetting parenthesis somewhere. Enter it like this: (sqrt(2) + 2 - sqrt(6)) / 4
I would just like to clarify that *I* was the one who was right. :D
 

thegreatjeff

Senior member
Jun 14, 2000
914
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0
I would just like to clarify that <STRONG>*I*</STRONG> was the one who was right. :D
Yep, you was...thanx a ton! =) (I just hope my math teacher isn't like "ok u got that right, but u didn't do the problem the way I wanted so u failed that one" he has a tendency to do that...grrr)

 

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