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

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ZeHanzBest ResponseYou've already chosen the best response.2
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 4c in your equation is positive, (which means c <4) all the points on the circle with midpoint (0,0) and radius sqrt(4c) 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

mathlillyBest ResponseYou've already chosen the best response.0
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

ZeHanzBest ResponseYou've already chosen the best response.2
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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