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Simplify math course work

  • #1

Homework Statement



Simplify [PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP249819e8ec729ig8433b00002e0fh793haeag630?MSPStoreType=image/gif&s=15&w=137&h=59 [Broken] 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...
 
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Answers and Replies

  • #2
Mentallic
Homework Helper
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I can't see what you posted. Can you write it here?
 
  • #4
Mentallic
Homework Helper
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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?
 
  • #5


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. ><
 
  • #6
Mentallic
Homework Helper
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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...?
 
  • #7


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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