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Thread: Problem based on square root

  1. #1
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    Problem based on square root

    If $\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

    a) x=3
    b) x= $\displaystyle \infty$
    c) x=6
    d) x=9

    Please tell the answer with complete explaination
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by siddscool19 View Post
    If $\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+......}}}$, then

    a) x=3
    b) x= $\displaystyle \infty$
    c) x=6
    d) x=9

    Please tell the answer with complete explaination
    The answer is a. As you take the square root of 6 infinitely and continuously add it to itself, you will slowly approach 1/2 of 6 which is three. You will get closer and closer to three, but the repeating square root prohibits is from getting there.
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  3. #3
    o_O
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    Let: $\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$

    Now, note that if you square it and move +6 to the LHS: $\displaystyle x^2 = 6 + \sqrt{6+\sqrt{6+ \cdots}} \ \ \Leftrightarrow \ \ x^2 - 6 = \sqrt{6+\sqrt{6+ \cdots}}$

    But the RHS is $\displaystyle x$ again!

    Thus, we see that $\displaystyle x$ is the positive solution to the quadratic: $\displaystyle x^2 - x - 6 = 0$

    which is simply 3.
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  4. #4
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    Quote Originally Posted by o_O View Post
    Let: $\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$

    Now, note that if you square it and move +6 to the LHS: $\displaystyle x^2 = 6 + \sqrt{6+\sqrt{6+ \cdots}} \ \ \Leftrightarrow \ \ x^2 - 6 = \sqrt{6+\sqrt{6+ \cdots}}$

    But the RHS is $\displaystyle x$ again!

    Thus, we see that $\displaystyle x$ is the positive solution to the quadratic: $\displaystyle x^2 - x - 6 = 0$

    which is simply 3.
    $\displaystyle x = \sqrt{6+\sqrt{6+\sqrt{6+ \cdots}}} > 0$
    $\displaystyle x=\sqrt{6+x}$
    $\displaystyle x^2=6+x$
    x=3 or -2

    ans=(a)3
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