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6

try with: Solve[a*x^2 + b*x + c == 0, x]


4

In Mathematica 10: ==Hello I would like to solve the quadratic equation please


4

For an exact solution: f[k_, x_] = -k^x + x^2; soln = Reduce[{f[k, x] == 0, D[f[k, x], x] == 0, x > 0}, {k, x}, Reals] // ToRules {k -> E^(2/E), x -> E} f[k, x] /. soln 0 k /. soln // N[#, 17] & 2.0870652286345330 Plot[f[k /. soln, x], {x, 2, 3}]


3

You can't solve that monster algebraically. One could create a function that takes all the parameters and numerically finds the velocity. u0is a starting point solveU[u0_, {deltaP_, L_, rho_, d_, eps_, mu_}] := u /. FindRoot[deltaP == (1.52648*10^21 L u^2 rho)/( d ((3.53972*10^19 u^2 eps^2 rho^2)/mu^2 - ( 5.7107*10^20 u ...


3

Alternatively, use FindRoot FindRoot[Integrate[ SquareWave[{0.2, 0}, ((x - 2.5)/10)], {x, 0 + a, 10 - a}] == 0.95, {a, .5}] {a -> 0.125}


3

Solve[{Integrate[ SquareWave[{2/10, 0}, ((x - 25/10)/10)], {x, a, 10 - a}, Assumptions -> 0 < a < 1] == 95/100, 0 <= a <= 1}, a, Reals] (* {{a -> 1/8}} *)


3

Having just one root requires eq1 = -k^x + x^2 == 0; eq2 = D[-k^x + x^2, x] == 0 (* 2*x - k^x*Log[k] *) FindRoot can solve for x and k simultaneously. FindRoot[{eq1, eq2}, {{x, 2.5}, {k, 2}}] (* {x -> 2.71828, k -> 2.08707} *) Plot[(-k^x + x^2) /. %[[2]], {x, 2, 3}]


2

First off: you have a lot of unnecessary bracket pairs. Your equation should be written acc == (d + d (-1 + 1/fpr) (-1 + 1/precision))/ (d + d (-1 + 1/precision) + d (-1 + 1/fpr) (-1 + 1/precision) + d (-1 + 1/tpr)) Second, since d can be eliminated from your equation, you can't get a solution in terms of d. Eliminate[ acc == (d + d ...


2

Restricting the domain to Reals will speed up the solution. NSolve[{ rp t1 - r (t1p - t2p) t3p + rp t4 - r (2 t1p + t2p) t4p == 0, t1p (r + t3) - rp t1 t3p - 2 t1p t4 - 2 rp t1 t4p == 0, -t2p t3 + rp t2 t3p - t2p t4 - rp t2 t4p == 0, -rp (t1p - t2p) t3 + (t1 - t2) t3p == 0, -rp (2 t1p + t2p) t4 + (r - 2 t1 - t2) t4p == 0, r^2 + rp^2 == 1, t1^2 ...


2

This expression can be simplified substantially as follows. Beginning with ((a - 1) (1 + d x^2))/(a (Exp[b x + c] - 1)) - e multiply the expression by its denominator and Simplify Numerator[Together[%]] (* -1 + a + a*e - a*e*E^(c + b*x) - d*x^2 + a*d*x^2 *) Collect[b^2/(d (a - 1)) %, x^2, Simplify] (* (b^2*(-1 + a*(1 + e - e*E^(c + b*x))))/((-1 + a)*d) ...


2

gradient[g_, vars_] := Table[D[g@@vars, vars[[j]]], {j, 1, Length[vars]}] system1[lstConst_, vars_] := Join[ Join@@ Table[gradient[lstConst[[j]], vars], {j, 1, Length[lstConst]}], Table[lstConst[[j]]@@vars,{j,1,Length[lstConst]}]]; system2[f_, lstConst_, vars_, lambda_] := Join[ gradient[f, vars] - Sum[ lambda[[j]]*gradient[lstConst[[j]], vars], ...


2

There's a misunderstanding here. The third "dom" argument is not simply a set over which we solve the equation. There are only a few choices that can be used for the domain argument, and they have very specific effects on how Solve works. An example from the documentation: If dom is Reals, or a subset such as Integers or Rationals, then all constants ...


1

{Solve [ x^2 == 1, {x}], Solve [ x^2 == 1 && x > 0, {x}]} (* {{{x -> -1}, {x -> 1}}, {{x -> 1}}}*)


1

So I think I found a way following this post: Find Roots in Do loop I added these commands: t = List[0, 0.5, 1, 1.5, 2] Table[FindRoot[{e1, e2, e3, e4} /. dat, {{w, 0.5}, {r, 2}, {p2, 0.8}, {pw, 1.8}}], {a, t}] And it does calculate FindRoot over the different values of a.


1

Just use Solve : Solve[ -11 x1 - 20 x2 - 9 x3 + 8 x4 - 9 x5 - 14 x6 - 9 x7 - 11 x8 + 3 x9 == 19 x1 + 19 x2 + 20 x3 - 12 x4 + 2 x5 - 13 x6 - 17 x7 + 10 x8 + 17 x9 == -14 15 x1 + 2 x2 + 17 x3 + 15 x4 - 20 x5 - 10 x6 - x7 + 9 x8 + 8 x9 == -6 13 x1 + 13 x2 - 8 x3 + 5 x4 - 8 x5 - 12 x6 + 11 x7 - 19 x8 + 11 x9 == -17 - 18 x1 - 5 x2 - 18 x3 - 7 ...


1

Use FindRoot: FindRoot[Sqrt[(2*m x b^2)/h^2]*Tan[Sqrt[(2*m x b^2)/h^2]] == Sqrt[(2*m*(V - x)*b^2)/h^2], {x, 5*10^-20}] (*Output*) {x -> 4.86164*10^-20}



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