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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: a very general maths ques
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[Post New]1 Jul 2008 19:54:57 IST
    Subject: a very general maths ques
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shreyaarya (92)

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if x = 9 - 4root5


find rootx + 1/rootx


the options are:


2


4


6


9


i think none of them is coorect.plzzz post ur solns.
    
[Post New]1 Jul 2008 21:00:29 IST
    Subject: Re:a very general maths ques
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vibhu.oct (466)

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x=9-4root5


=> 1/x = 9+4root5


[rootx + 1/rootx]2  => x + 1/x + 2  =>9-4root5+9+4root5+2  =>20


So, rootx + 1/rootx = root20 = 2root5.


the ans. is not in the choices.


Rate, if useful. Please correct , if I am wrong.


 
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[Post New]1 Jul 2008 22:08:30 IST
    Subject: Re:a very general maths ques
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rudra.panda (2507)

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x=9-4\sqrt{5}\\\\\frac{1}{x}=9+4\sqrt{5}\\\\(\sqrt{x}+\frac{1}{\sqrt{x}})^2=x+\frac{1}{x}+2\\\\\Rightarrow(\sqrt{x}+\frac{1}{\sqrt{x}})^2=9-4\sqrt{5}+9+4\sqrt{5}+2\\\\\Rightarrow(\sqrt{x}+\frac{1}{\sqrt{x}})^2=18+2=20\\\\\Rightarrow\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{20}\\\\\Rightarrow\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{5}


\text{Hence No option is correct}.
God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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