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8

Per @J.M. 's suggestion: gb = GroebnerBasis[{Sqrt[3 y] + Sqrt[x^2 + 2 x] - x - x Sqrt[2 + 9 y^2], y (Sqrt[x y - 2 y^2] + Sqrt[4 y^2 - x y]) - (2 x^2 - 5 x y - y^2)}, {x, y}]; eq = Thread[gb == ConstantArray[0, Length[gb]]]; Solve[eq, {x, y}, Reals] {{x -> 0, y -> 0}, {x -> 1, y -> 1/3}}


4

Try something like x /. Solve[Sum[RandomChoice[{-1, 1}] x^k, {k, 0, 3}] == 0, x, Reals] instead of your Roots expression. Use NSolve if you want the numerical value of the roots, rather than a symbolic representation. For instance: t = Table[x /. NSolve[Sum[RandomChoice[{-1, 1}] x^k, {k, 0, 3}] == 0, x, Reals], 4500]; realroots = DeleteDuplicates@...


4

For your 1st problem, assume the following. r is radius of small circles. s is the distance between the center of the large circle and the center of any of small circles. R is the radius of large circle. r == 15 is given; r == s Sin[60 °] and R == r + s by inspection, so With[{r = 15}, Solve[Eliminate[{r == s Sin[60 °], R == r + s}, s], R]][[1,1]] ...


3

This is just to illustrate for the first problem (calculation and visualization). Module[{tg = SSSTriangle[30, 30, 30], c, ans}, c = RegionCentroid[tg]; ans = EuclideanDistance[c, tg[[1, 1]]] + 15; Graphics[{Circle[c, ans], LightGray, Disk[#, 15] & /@ tg[[1]], Red, PointSize[0.02], Point[tg[[1]]], EdgeForm[Black], FaceForm[None], Polygon[tg[[...


3

Example Code pts = Solve[y - 2 x^2 + 3/2 == 0 && {x, y} \[Element] Circle[{0, 0}, 1],{x,y}]; parabola = ContourPlot[{y - 2 x^2 + 3/2 == 0}, {x, -1.5, 1.5}, {y, -1.5, 1.5}]; intersections = {Red, PointSize[Medium], Point[{x, y} /. pts]}; Show[{parabola, Graphics[{Circle[{0, 0}, 1], intersections}]}] Output Reference Curve Intersection Show ...


3

If you accept that the general solution can be constructed as an (infinite) Fourier series, $$\sum_{m=-\infty}^{\infty}\phi_m(r) \exp(i m \theta),$$ then you can obtain the expression for $\phi_m(r)$ as follows: eqn = 0 == Simplify@ Laplacian[ Laplacian[ ϕ[r] Exp[I m θ], {r, θ}, "Polar"], {r, θ}, "Polar"]; DSolve[eqn, ϕ, r] (* ==> {{ϕ -> ...


2

Seems to be a case of ill conditioning of the input system. If I redo using exact input and set NSolve to work on high precision then I get a plausible outcome. c1 = 50; c2 = 3/2; c3 = 2/5; c4 = 1/2; c5 = 8; c6 = 16/5; k1 = 37*10^(-17); k2 = 83*10^(-8); k3 = 87*10^(-12); k4 = 32*10^(-6); k5 = 86*10^(-32); eqs = {f1*f2 - k1*x1^2, f1*x3 - k2*x2*x1, f1*(c6 ...


2

In general one should not expect to obtain a general symbolic solution (a function x[a,b]) to the given equation since there are two independent variables, see e.g. Solve symbolically a transcendental trigonometric equation and plot its solutions for certain aspects regarding transcendental equations. To get an idea how the solution depends on parameters a ...


2

Perhaps this! It works if you only need the plot and not the values, or the function. That would require more work. ContourPlot3D[ Evaluate[-x + (Log[f2[x]] - Log[f1[x]] + Log[b] - Log[a])/((1 - a) + (1 - 1/f1[x]) - (1 - b) + (1 - 1/f2[x]))] , {a, 0, 1}, {b, 0, 1}, {x, 0, 1} , AxesLabel -> {"a", "b", "x"} , Contours -> {0} , Mesh -> None ...


1

ubpdqn beat me to it and his looks better. You can have no knowledge about the problem at all and still get Mma to get the answer, just use it like Geogebra on steroids. t = SSSTriangle[30, 30, 30]; cent = RegionCentroid[t]; Maximize[y, {x, y} \[Element] Circle[cent, 15]]; big = Circle[cent, 15 + 15 Sqrt[3] - 5 Sqrt[3]]; Graphics[{t, Circle[{0, 0}, 15], ...


1

To find the exact values for min and max roots roots = DeleteDuplicates@ Flatten[x /. Solve[# == 0, x, Reals] & /@ (Tuples[{-1, 1}, {4}].{1, x, x^2, x^3})] // SortBy[#, N] &; {min, max} = roots[[{1, -1}]] // ToRadicals The approximate numeric values are as shown by @MarcoB {min, max} // N (* {-1.83929, 1.83929} *)


1

This reminds me of NSolve finds real-valued results in version 9, but not in version 10, in which you can use any of the Method settings from Methods for NSolve "EndomorphismMatrix" "CompanionMatrix" "Legacy" "Aberth" "JenkinsTraub" or even a nonexistent method "Foo": NSolve[eqs, {x1, x2, x3, x4}, Reals, Method -> "Foo"] (* {{x1 -> 23.2915, x2 -&...


1

In M10, you will get the answer given by Feyre. Solve[c1 && c2 && c3 && c4] {{b1 -> 0.551683 - 0.197769 I, b2 -> 0.163159 - 0.453896 I, b3 -> 0.29314 - 0.549955 I, b4 -> -1.81831*10^-10 + 9.53055*10^-10 I, r1 -> -0.336636 - 0.847225 I, rIm1 -> 0.895327 - 1.06646 I, t1 -> -0.245692 + 0.329425 I, tIm1 -> -...



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