Simplify

[Michael_Light]

1. The problem statement, all variables and given/known data

Simplify http://www4c.wolframalpha.com/Calculate/MSP/MSP249819e8ec729ig8433b00002e0fh793haeag630?MSPSto reType=image/gif&s=15&w=137&h=59 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...

[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 (http://img190.imageshack.us/i/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! ^^