Simplify Math Course Work: How to Rationalize Denominators with Conjugates

[Michael_Light]

Homework Statement

Simplify [PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP249819e8ec729ig8433b00002e0fh793haeag630?MSPStoreType=image/gif&s=15&w=137&h=59 Please kindly show the working steps as well. ^^ Thanks.

Homework Equations

The Attempt at a Solution

I tried but i have no idea on how to start...

 

[Mentallic]

I can't see what you posted. Can you write it here?

 

[Michael_Light]

I can't see what you posted. Can you write it here?

Sorry, here it goes: http://img190.imageshack.us/img190/5519/msp227119e8f3eb0185c51e.gif

 

[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?

 

[Michael_Light]

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]

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...?

 

[Michael_Light]

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! ^^