# Simplify

by Michael_Light
Tags: simplify
 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...
 HW Helper P: 3,436 I can't see what you posted. Can you write it here?
P: 117
 Quote by Mentallic I can't see what you posted. Can you write it here?
Sorry, here it goes:

HW Helper
P: 3,436

## Simplify

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?
P: 117
 Quote by Mentallic 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. ><
 HW Helper P: 3,436 If you have something like $$\frac{1}{1+\sqrt{x}}$$ then you can get rid of any roots in the denominator (bottom part of the fraction) by multiplying by the conjugate $$1-\sqrt{x}$$ Basically, the conjugate of a+b is a-b. When you multiply $$1+\sqrt{x}$$ by $$1-\sqrt{x}$$ you get $$1-x$$. When you multiply a-b by a+b you get $$a^2-b^2$$ 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 $$\frac{1}{1+\sqrt{x}}=\frac{(1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})}=\frac{1-\sqrt{x}}{1-x}$$ That is what you call rationalizing the denominator. Now, for your question, the conjugate of $$\sqrt{1-x^2}+\sqrt{1+x^2}$$ will be...?
P: 117
 Quote by Mentallic If you have something like $$\frac{1}{1+\sqrt{x}}$$ then you can get rid of any roots in the denominator (bottom part of the fraction) by multiplying by the conjugate $$1-\sqrt{x}$$ Basically, the conjugate of a+b is a-b. When you multiply $$1+\sqrt{x}$$ by $$1-\sqrt{x}$$ you get $$1-x$$. When you multiply a-b by a+b you get $$a^2-b^2$$ 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 $$\frac{1}{1+\sqrt{x}}=\frac{(1-\sqrt{x})}{(1+\sqrt{x})(1-\sqrt{x})}=\frac{1-\sqrt{x}}{1-x}$$ That is what you call rationalizing the denominator. Now, for your question, the conjugate of $$\sqrt{1-x^2}+\sqrt{1+x^2}$$ 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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