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    This question is C1 January 2006, Q10 part (d), for reference.

    I got to k^(2)<12, and then k<=+/-2root3. I thought I should leave the answer as k<2root3, as k<-2root3 is inclusive in k<2root3.

    The answer is -2root3<k<2root3, and I am not sure why. Could somebody explain please?

    Thank you
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    Thing about the graph of  y=x^2 -12 , to satisfy the inequality (replacing k with x) the curve must be below the x axis so that y<0.

    You cannot just square root it when there is an inequality. By your logic -10 should satisfy the inequality, but does it in reality? No.
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    It is k>-2root3 as the square of (-2root3) is 12. If we make it more negative then k^2 will be greater than 12.


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    I loved your explanation (http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2-12)
    It is like finding the roots of X^2-12=0
    Generally I try remembering that x^{2}\leq a\Rightarrow-\sqrt{a}\leq x\leq\sqrt{a}

    (Original post by B_9710)
    Thing about the graph of  y=x^2 -12 , to satisfy the inequality (replacing k with x) the curve must be below the x axis so that y<0.

    You cannot just square root it when there is an inequality. By your logic -10 should satisfy the inequality, but does it in reality? No.
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    (Original post by jamb97)
    ...
    In addition to the above, here's a pretty picture.



    The two intersections of the curve and the line are at \pm 2\sqrt{3} - I'm sure you can tell which is which. Now for what values of k is the curve under the or less than the line?


    (Original post by depymak)
    x^{2}\leq a\Rightarrow-\sqrt{a}\leq x\leq\sqrt{a}
    To see why this is true think about x^2 \leq a \Rightarrow \sqrt{x^2} \leq \sqrt{a} since the square root is monotone increasing function then |x| &lt; \sqrt{a}.
 
 
 
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