# Math Help - Quadratic Question

Hi,

This is probably a relatively easy question, but I've never done this before, and I'm not sure how to go about it.

For the equation $3x^2+ax+4=0$ find the smallest possible value of $a$ such that the equation will have two distinct rational solutions.

*Edit - Never-mind. I just figured it out

3. Hello, Stroodle!

For the equation $3x^2+ax+4\:=\:0$, find the smallest possible value of $a$
such that the equation will have two distinct rational solutions.

*Edit - Never-mind. I just figured it out
Good for you!

But is this a trick question?
I noted that it didn't say smallest positive value of $a$.

Using the Quadratic Formula: . $x \;=\;\frac{-a \pm\sqrt{a^2-48}}{6}$

For rational roots, the discriminant $(a^2-48)$ must be a square.

This is true for: . $a \;=\;\pm7,\:\pm8,\:\pm13$

Then the smallest (least) value of $a$ is: . $a \:=\:-13$

4. Haha. Well spotted Soroban!

Actually the question did ask for the smallest positive value of $a$, but in my tired & vague state of mind I accidentally failed to copy it down...