How do I determine the values of $k$ for which $3x^2 + kx+12 = 0$ has no real solutions, $1$ real solution, and $2$ real solutions?

Question

I know that a point has to be $(0,12)$ just because of the $C$ term always being the y-intercept.

Please just tell me how to approach the problem/give tips, so that you are not doing my homework for me.

Answer 1

hint: $\triangle = b^2 - 4ac = k^2 - 4(3)(12) = k^2 -144$. If $\triangle = 0$, it has $1$ real solution, $ < 0$ it has no real solution, and $ > 0$ it has $2$ real solutions.

Answer 2

By the distributive property the equation is equivalently:

$$0=3\left(x^2+\frac{k}{3}x\right)+12$$

Subtracting $12$ from both sides of the equation.

$$-12=3\left(x^2+\frac{k}{3}x\right)$$

Dividing by $3$ on both sides of the equation.

$$-4=x^2+\frac{k}{3}x$$

Using the fact that $\left(x+\frac{y}{2}\right)^2=x^2+2\left(x\right)\left(\frac{y}{2}\right)+\frac{y^2}{4}$:

$$-4=\left(x+\frac{k}{6}\right)^2-\frac{k^2}{36}$$

$$\frac{k^2}{36}-4=\left(x+\frac{k}{6}\right)^2$$

Note now that if $u^2=c$ then $c=0$ gives one solution $0$, $c>0$ gives $2$ solutions $\pm \sqrt{c}$, and $c<0$ gives no real solutions.

With a little more thinking we can see that If we want $1$ real root we need:

$$\frac{k^2}{36}-4=0$$

If we want $2$ real solutions we need:

$$\frac{k^2}{36}-4>0$$

And finally no real solutions:

$$\frac{k^2}{36}-4<0$$