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    Okay, if you have (x^2+16)^{.5}, does that imply a +- of the root itself, or just the positive root? I ask this because I've been asked to sketch such graph and the solution only includes the sketch above the x-axis. The question did not restrict the range or anything.
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    (Original post by ViralRiver)
    Okay, if you have (x^2+16)^.5, does that imply a +- of the root itself, or just the positive root? I ask this because I've been asked to sketch such graph and the solution only includes the sketch above the x-axis. The question did not restrict the range or anything.
    That's just an expression? Did you mean to say "sketch \sqrt{x^2+16}? If so, you only sketch the positive part because the RHS is clearly just positive.

    The confusion you're having is that if you're just given something as a surd, it is always positive. Whereas, if you had y^2=x^2+16 and then took the square root of both sides, you would have y=\pm \sqrt{x^2+16} and would thus have to sketch both the positive and negative parts of the graph.
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    (Original post by ViralRiver)
    Okay, if you have (x^2+16)^{.5}, does that imply a +- of the root itself, or just the positive root? I ask this because I've been asked to sketch such graph and the solution only includes the sketch above the x-axis. The question did not restrict the range or anything.
    Check this link:

    http://www.wolframalpha.com/input/?i=%28x^2%2B16%29^-5

    You will find the graph there as well.
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    Okay, so if you had y^2=x and you rewrote it as y=x^{\frac{1}{2}}, then clearly you're ignoring one of the roots. Is there anyway to get round that with indices, and avoiding the square root symbol?
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    (Original post by ViralRiver)
    Okay, so if you had y^2=x and you rewrote it as y=x^{\frac{1}{2}}, then clearly you're ignoring one of the roots. Is there anyway to get round that with indices, and avoiding the square root symbol?
    If you're solving that top equation for y in terms of x, then yes you are ignoring the negative root.

    It's not a case of getting around it. The only way you can discard it is if you define y>0.

    If I'm honest, I'm not sure what your query is. What do you mean by "get around it"?
 
 
 
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