Why not formulate Quantum Mechanics using Lagrangians? As the title implies, why is it that the most common formalisms we use in quantum mechanics prefer to describe systems in the terms of a Hamiltionian instead of a Lagrangian? 
Is there some convenience to defining our systems one way over the other? Are there cases I'm not aware of where Lagrangian formalism is preferred? 
 A: It's because they're based on the historical approach: Schroedinger's equation.
Schroedinger's equation was discovered on its own before we knew about canonical quantization.  Dirac came up with the canonical quantization rules which re-wrote (and generalized) Schroedinger's equation into the familiar one we have today, $\hat{H} \psi = i \dot{\psi}$.
That said, there is an approach which uses the action (and thus the Lagrangian or Lagrangian density) due to Feynman: The path integral approach.  This approach has as its biggest advantage the ability to be reconciled with Special Relativity, which proved much too difficult a task for extensions of the Schroedinger equation (the Dirac Equation was the most successful attempt, but wasn't general enough to describe some phenomena).
This is what is used in the most advanced quantum physics such as quantum field theory, quantum electrodynamics being the best example.  But unless you're interested in high-energy particle physics or really advanced condensed matter physics, the traditional quantum mechanics is sufficient.
A: I don't have an answer for why there is no simple Lagrangian formulation, but I can explain some of why a Hamiltonian one is easy. Part of the way to go from Classical Mechanics to Quantum is by replacing Poisson brackets with commutators, and observables with operators on Hilbert space and their expectation values. So the equation 
$\frac{d}{dt} f(q, p, t) = \left\{ f,H \right\} + \frac{\partial f}{\partial t} $
becomes the quantum
$\frac{d}{dt} \langle f \rangle = -i \langle\left[f,H \right]\rangle + \langle \frac{\partial f}{\partial t} \rangle.$
So the Hamiltonian is convenient because it gives the time evolution of operators, states, and expectation values directly. Also, because the Hamiltonian is a conserved quantity, stationary states (i.e. those that do not evolve in time) will be eigenvectors of the Hamiltonian, and eigenvalue problems are easy. 
A: I can think of several reasons for why using Hamiltonians is preferred, but the most important, I'ld say, is that you need to use path integral formalism in order to formulate (non relativistic) QM in terms of the Lagrangian, which, for an undergrad course, is a bit of an overkill.
Also, many of the most renowned equations in QM like, say, the Schrödinger Equation, use the Hamiltonian:
$\hat H \Psi=\hat E \Psi$
Thus, although it is possible, why change it? It would be quite a pain to do so.
For what I understand, however, modern QM relies heavily on both the Hamiltonian and the Lagrangian formalism.
Hope it helped!
