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Simplify math course work

by Michael_Light
Tags: simplify
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Michael_Light
#1
Feb12-11, 09:16 PM
P: 117
1. The problem statement, all variables and given/known data

Simplify Please kindly show the working steps as well. ^^ Thanks.

2. Relevant equations



3. The attempt at a solution

I tried but i have no idea on how to start...
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Mentallic
#2
Feb12-11, 11:16 PM
HW Helper
P: 3,562
I can't see what you posted. Can you write it here?
Michael_Light
#3
Feb12-11, 11:55 PM
P: 117
Quote Quote by Mentallic View Post
I can't see what you posted. Can you write it here?
Sorry, here it goes:

Mentallic
#4
Feb13-11, 12:07 AM
HW Helper
P: 3,562
Simplify math course work

Ok, by simplification I assume they mean rationalizing the denominator (getting rid of square roots in the denominator). Do you know about multiplying by the conjugate?
Michael_Light
#5
Feb13-11, 12:26 AM
P: 117
Quote Quote by Mentallic View Post
Ok, by simplification I assume they mean rationalizing the denominator (getting rid of square roots in the denominator). Do you know about multiplying by the conjugate?
I guess i don't know how to multiply by conjugate since i don't even know what is a conjugate. ><
Mentallic
#6
Feb13-11, 01:13 AM
HW Helper
P: 3,562
If you have something like [tex]\frac{1}{1+\sqrt{x}}[/tex] then you can get rid of any roots in the denominator (bottom part of the fraction) by multiplying by the conjugate [tex]1-\sqrt{x}[/tex]
Basically, the conjugate of a+b is a-b. When you multiply [tex]1+\sqrt{x}[/tex] by [tex]1-\sqrt{x}[/tex] you get [tex]1-x[/tex]. When you multiply a-b by a+b you get [tex]a^2-b^2[/tex] so you can see that if a and b are square roots, the square roots will vanish in the denominator. So multiplying by the top and the bottom will give you [tex]\frac{1}{1+\sqrt{x}}=\frac{(1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})}=\frac{1-\sqrt{x}}{1-x}[/tex]
That is what you call rationalizing the denominator.

Now, for your question, the conjugate of [tex]\sqrt{1-x^2}+\sqrt{1+x^2}[/tex] will be...?
Michael_Light
#7
Feb13-11, 02:10 AM
P: 117
Quote Quote by Mentallic View Post
If you have something like [tex]\frac{1}{1+\sqrt{x}}[/tex] then you can get rid of any roots in the denominator (bottom part of the fraction) by multiplying by the conjugate [tex]1-\sqrt{x}[/tex]
Basically, the conjugate of a+b is a-b. When you multiply [tex]1+\sqrt{x}[/tex] by [tex]1-\sqrt{x}[/tex] you get [tex]1-x[/tex]. When you multiply a-b by a+b you get [tex]a^2-b^2[/tex] so you can see that if a and b are square roots, the square roots will vanish in the denominator. So multiplying by the top and the bottom will give you [tex]\frac{1}{1+\sqrt{x}}=\frac{(1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})}=\frac{1-\sqrt{x}}{1-x}[/tex]
That is what you call rationalizing the denominator.

Now, for your question, the conjugate of [tex]\sqrt{1-x^2}+\sqrt{1+x^2}[/tex] will be...?
Oo.. i don't know that is called conjugate. >< By rationalizing the denominator with its conjugate, now i managed to solve the question! Thanks! ^^


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