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Math Help - Root-simple,but...

  1. #1
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    Root-simple,but...

    Please;

    how do I get \sqrt{x+1}=x\sqrt{}1/x^2-1
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  2. #2
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    Re: Root-simple,but...

    Quote Originally Posted by marijakopljar View Post
    Please;

    how do I get \sqrt{x+1}=x\sqrt{}1/x^2-1
    Did you mean  \sqrt{x+1} = x \sqrt{\frac{1}{x^2-1}}

    Kalyan
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  3. #3
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    Re: Root-simple,but...

    How do you get this equation? How should we know?

    Perhaps you're really asking how do you SOLVE this equation?
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  4. #4
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    Re: Root-simple,but...

    Quote Originally Posted by marijakopljar View Post
    Please;

    how do I get \sqrt{x+1}=x\sqrt{}1/x^2-1
    What you wrote is the same as \sqrt{x+1}= x/x^2- 1= (1/x)- 1. What you perhaps intended was \sqrt{x+ 1}= x\sqrt{\frac{1}{x^2- 1}}= \frac{x}{\sqrt{x^2- 1}}
    But "how do I get" an equation usually means "how do I show that these are the same for all x". And what you have isn't. For example if x=3, the left side (of both of those) is \sqrt{4}= 2 while the right side of the last is \frac{3}{\sqrt{9- 1}}= \frac{3\sqrt{2}}{4}, NOT 2. For the first, it is 1/3- 1= -2/3.

    If you mean "how do I solve \sqrt{x+ 1}= x\sqrt{\frac{1}{x^2- 1}?} the simplest thing to do is get rid of the square roots by squaring both sides: x+ 1= \frac{x^2}{x^2- 1} which, mutiplying both sides by x^2- 1 gives (x+ 1)(x^2- 1)= x^3+ x^2- x- 1= x^2 so you have to solve the cubic equation x^3- x- 1= 0. However, that has no rational number roots so I doubt if that is what you mean.
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  5. #5
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    Re: Root-simple,but...

    There is some doubt in the explanation. What do you want to do??
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