In the example I used the same matrices as Markiyan:

with(LinearAlgebra):

A:=Matrix([[-93, -72], [-76, -2]]):

C:=Matrix([[-98, 57], [-77, 27]]):

E:=Matrix([[-38, 87], [-18, 33]]):

Id:=IdentityMatrix(2):

fsolve(Determinant(A-lambda*C.Id-lambda^2*E.Id), lambda,complex);

           -11.36829371, -.3123524009-1.257844764*I, -.3123524009+1.257844764*I, .8872292854

The next version works:

restart;
for a from 0 to 10 do
A := Matrix([[a, 0], [0, 0]]):
if LinearAlgebra[Equal](A^2, A) then print(a, A) end if;
end do:

r:=[a,b,c]:

[seq(lambda-r[i]^2, i=1..3)];

I assumed that the sign of inequality is >= 0. The following simple procedure solves your problem.

P:=proc(alpha, beta)

if beta=-1 then error "0 in denominator" fi;

if alpha=0 then {k>0} else allvalues(solve({alpha*beta*(1-alpha)*k^(alpha-1)/(1+beta)-1-(2*alpha-1)*alpha*(1-alpha)*beta*k^(2*(alpha-1))/(1+beta)+alpha^2*k^(alpha-1)>=0, k>=0}, k))  fi;

end proc:

Examples:

P(1/2, 3),  P(7, 1/2),  P(3, 5),  P(1, 3);

M. Hirnyk!
Of course, the eliminate command is not always able to find an explicit equation relating B and A. In this case, the problem can be solved numerically by the following scheme (the simplest way):

1) Construct the plot of the function  B(t)  and on it we find the range in which the function is monotonic.
2) For each value of  B  through some step find the corresponding value of t .

3) Substitute these  t in A (t) .
4) Using these lists we build the plot A (B) .

Solution for your example:

plot(t^5-sin(t)-2, t=0..2, labels=[t,B]);

T:=[seq(fsolve(B = t^5-sin(t)-2, t=0.63..2), B=-2.4..29, 0.1)];

X:=[seq(B, B=-2.4..29, 0.1)];
Y:=[seq(eval( t^3-exp(t)), t in T)];

plot(X, Y, labels=[B,A]);

You have 2 functions  A=A(t)  and B=B(t) .To plot A in terms of B you should to eliminate the variable  t . 

An example:

F := eliminate({A = t^3-t, B = t^3-t^2}, t);

plots[implicitplot](op(F[2]), B = -1 .. 1, A = -1 .. 1, scaling = constrained, numpoints = 50000);

If I understand your question, you need to build a point plot of the function  x->x/(x+1)  for a sets of points  0.5, 0.9, 0.99, 0.999  and  1.001, 1.01, 1.1, 1.5  to illustrate the concept of limit of a function when the variable tends to 1. This can be done as follows

X1, X2 := [.5, .9, .99, .999], [1.001, 1.01, 1.1, 1.5]:

f := x->x/(x+1):

Y1, Y2 := map(f, X1), map(f, X2):

plots[display](plot(X1, Y1, style = point, symbol = solidcircle, color = blue, symbolsize = 8), plot(X2, Y2, style = point, symbol = solidcircle, color = red, symbolsize = 8), view = [0 .. 2, 0 .. 1], scaling = constrained);

restart;

f := -k*x1(t)+k*(x2(t)-x1(t))-lambda*(diff(x1(t), t)) ;

g := -k*(x2(t)-x1(t));

eq1 := m*(diff(x1(t), t, t))-f = 0;

eq2 := m*(diff(x2(t), t, t))-g = 0;

fcns := {x1(t), x2(t)};

ICS := {x1(0) = 1, x2(0) = 0, D(x1)(0) = 0, D(x2)(0) = 0};

sys := {eq1, eq2}; m := 1; k := 1; lambda := 1;

sysdiff := sys union ICS;

Sol := dsolve(sysdiff, fcns, numeric);

plots[odeplot](Sol, [[t,x1(t)],[t,x2(t)]], t=0..30, color=[red, blue], numpoints=5000, thickness=2);

restart :

f := -k*x1(t)+k2*(x2(t)-x1(t))-lambda*(diff(x1(t), t)):

g := -k2*(x2(t)-x1(t)): fcns := x1(t), x2(t):

eq1 := m*(diff(x1(t), t, t))-f = 0:

eq2 := m*(diff(x1(t), t, t))-g = 0:

ICS := {x1(0) = 1, x2(0) = 0, D(x1)(0) = 0}:

sys := {eq1, eq2}:

m := 1: k2 := 1: k1 := 1: lambda := 1: k:=1:

sysdiff := `union`(sys, ICS):

Sol_N := dsolve(sysdiff, {fcns}, numeric):

plots[odeplot](Sol_N , [x1(t),x2(t)],0..100, numpoints=5000, thickness=2, view=[-0.5..0.5, -0.5..0.5]);

a[n] is `in` S;

plot([[4*cos(t), 3*sin(t), t=0..2*Pi], [2*cos(t), 3*sin(t), t=0..2*Pi]], scaling=constrained, color=[red, blue]);

f:=x->x^2+2*x-3:
D(f)(2);

                  6

Simplification.mw

restart;

L:=[]:

 for A  from .1 by .01 to 3 do

  for B  from .1 by .01 to 3 do

  l1:=(exp(B)-A)-1:

  l2:=(exp(B)-A/2)-1.5:

  if l1>0 and l2>0 then L:=[op(L), [A,B]]: break: end if:

  end do:

 end do:

L; 

In total there are 192 solutions:

with(combinat):

S:=permute(9): L:=[]:

for a in S do

if  a[9] = add(a[2*k-1]/a[2*k], k= 1..4) then L:=[op(L), a]: fi:

od:

nops(L);

                                               192

The first solution in the list  L  is Carl's solution, the last is

It is interesting that  there is no solution in which all the fractions are irreducible.