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why does x^2+y^2 = 4-c have no real solutions? (book example about functions of several variables and graphing functions)

  • one year ago
  • one year ago

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  1. ZeHanz
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    This equation could actually have real solutions. It depends on the value of c. The equation of a circle with midpoint (0,0) and radius r is: x²+y²=r² This means, as long as 4-c in your equation is positive, (which means c <4) all the points on the circle with midpoint (0,0) and radius sqrt(4-c) are (real) solutions. If c=4, then only the pair x=0, y=0 is a solution. If c>4, there are no real solutions.

    • one year ago
  2. mathlilly
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    That makes sense, but if I consider all solutions on one graph, would that make the valid ones, not valid? Does my question make sense?

    • one year ago
  3. ZeHanz
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    I'm not really sure what you mean, but if it is required that c can be any real value, then the conclusion remains: for some c there are solutions, for other there are none. For each of the values of c that are valid (c<4) you can draw graphs of these solutions. In the image you see the graphs of the solutions for c = {-5,-4,-3,...,3,4} The largest circle is the graph for c=-5, the origin is the solution when c=4.

    • one year ago
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