To piggyback off Bill's answer, one can just use CountRoots[]
on the characteristic polynomial of the given matrix pencil, if one only wishes to show that the eigenvalues are all real:
CountRoots[CharacteristicPolynomial[{B, A}, x], x]
4
One can then use RootIntervals[]
to find brackets for the roots:
RootIntervals[CharacteristicPolynomial[{B, A}, x], Reals]
{{{0, 0}, {0, 1}, {3, 4}, {4, 10}}, {{1}, {1}, {1}, {1}}}
Note that the root at $x=0$ was exactly isolated. The largest eigenvalue of the pencil would correspond to the last entry with the isolating interval $(4,10)$, which you can then give to Solve[]
:
Solve[CharacteristicPolynomial[{B, A}, x] == 0 && 4 < x < 10, x,
Cubics -> False, Quartics -> False]
{{x -> Root[-19440 + 76898 #1 - 28959 #1^2 + 2401 #1^3 &, 3]}}
Bob has already mentioned casus irreducibilis; to summarize, if you insist on a radical representation, then the use of a complex representation is (often) unavoidable, even if all the roots are real.