We were taught that Lagrange points are where the resultant gravitational intensity is zero. Does that mean the resultant gravitational attraction of an object kept there is zero? If that's the case how can it orbit, the resultant should be equal to the centripetal force right?
[Physics] What are Lagrange points
centrifugal forcedefinitionnewtonian-gravitynewtonian-mechanicsorbital-motion
Related Solutions
Yes the free body moves outward, but there are two critical things you have to know to interpret this statement correctly.
First, this is the effective potential, taking into account gravity and centrifugal force. It has this form because we went into the non-inertial frame co-rotating with the two masses. Mathematically, the potential is $$ \Phi_\mathrm{eff}(\vec{r}) = -G \left(\underbrace{\frac{M_1}{\lvert \vec{r}-\vec{r}_1 \rvert}}_\text{potential from mass 1} + \underbrace{\frac{M_2}{\lvert \vec{r}-\vec{r}_2 \rvert}}_\text{potential from mass 2} + \underbrace{\frac{M_1+M_2}{2\lvert \vec{r}_1-\vec{r}_2 \rvert^3} \lvert \vec{r} \rvert^2}_\text{centrifugal component}\right), $$ and it only decreases far away because of that last term.
Physically, this is because placing an object "at rest" in this frame corresponds to having it move with the same angular frequency as $M_1$ and $M_2$ about the center of mass. If you initialize an object $5\ \mathrm{AU}$ on a tangential path having the same angular velocity as the Earth, it will be moving too fast for a circular orbit at that distance, and so it will move away from the Sun.
This does not mean the object will go away forever, and that brings us to the second point, explained in Chay's response: Not all effective forces have been accounted for; in particular, the Coriolis force does not arise from $\Phi_\mathrm{eff}$. The Coriolis force depends on velocity, so it has no scalar potential depending solely on position, and so it is not included in the analysis so far. Once your test object starts moving in your rotating frame, it will experience a perpendicular deflection that will eventually force it to turn around.
There are several inaccuracies in your post.
First - you don't need equal mass (or one much smaller than the other) for a circular orbit: you need the least amount of energy for a given angular momentum. You can have unequal masses in circular orbit (about their center of gravity).
Second - while the force of gravity is pulling towards the focus of the ellipse, the direction of the orbit need not be perpendicular to the radial vector. When it is not, then part of the force will "slow down" or "speed up" the object in orbit, and part of it will "bend" the direction:
In this case, the red component is the one that is providing centripetal force, and the green component is accelerating the particle in orbit.
Best Answer
In the three-body problem, the Lagrange points are those points in space where two bodies with large mass (Earth and Sun), through the interaction of the respective gravitational force, allow a third body with a much lower mass to maintain a stable position relative to them.
In a planetary system it implies that a small object, such as a satellite or an asteroid, which shares the same orbit of a planet and positioned in a Lagrange point, will keep constant the distances between the major celestial bodies, the star and the planet with which shares the orbit.
For this to happen, the resultant of the gravitational accelerations imparted by the celestial bodies to the object must be exactly the centripetal acceleration necessary to keep the object in orbit at that particular distance from the largest celestial body, with the same angular velocity as the planet.
Out of curiosity: you can also verify using Lagrangian mechanics these five Lagrangian points.
Let be $(P-O)=\rho \vec{e}_\rho$ the distance between our third point and the mass centre of the system; $(T-O)=x_E \vec{e}_x$ the distance between the Earth and the CM and $(S-O)=x_S \vec{e}_x$ the distance between the Sun and the CM (very near to the centre of the Sun).
You will find that the system potential energy is $$U_{tot}=U_{grav}+U_{centr}=-G\frac{M_E\cdot m}{||P-T||}-G\frac{M_S\cdot m}{||P-S||}-\frac{1}{2}m\Omega^2\rho^2$$ If you derivate this energy potential respect to $\rho$ and $\theta$ you will find these points. And also, if you make the Hessian matrix you will find that 3 of them are points of instability $(L_1, L_2, L_3)$ and the other 2 are points of stability $(L_4, L_5)$.
I do not pretend that you understand perfectly these calculations but only the principal idea that is behind.
I hope that the first part of the answer could be helpful for you and maybe also the second one.