# Parabola and line intersection (basic algebra)

I am presented with these equations for a straight line and a parabola: $$y=2x-k$$
$$y=3x^2+2kx+5$$

with the goal of calculating the range of values for k such that the line and parabola do not intersect.

So far I’ve tried to find the turning point of the quadratic, though in terms of k. -k/3, $$(15-k^2)/3$$

. I think whatever k is, it must be less than the y value of the Quadratic’s turning point, but I’m not sure how to proceed from there.

Any help is greatly appreciated

• Do you know about discriminants? This will be helpful. Sep 21, 2019 at 19:38

$$3x^2+2kx+5=2x-k\iff3x^2+(2k-2)x+(5+k)=0$$

so the line and the parabola don't intersect iff the above quadratic equation has no real solutions, and this happens iff its discriminant is negative, so it must be:

$$\Delta=(2k-2)^2-4\cdot3\cdot(5+k)<0\iff4k^2-20k-56<0\iff$$

Can you continue from here?

• thank you so much! I truly pored over this problem like the plebeian I am in the world of math lol but could you please explain more about why it’s okay to set the original quadratic and linear functions equal to each other in the first place? I believe that I understand why it’s okay to assume the discriminant of the new quadratic <0, (in order to find the values of k that make it true) but I’m unsure of the rationale behind setting the original quadratic and linear function equal in the first place.sorry for being a thicko XD Sep 21, 2019 at 20:48
• @UbaidHassan If two functions $\;y=f(x),\,y=g(x)\;$ intersect at some point, then we can write down that point as $\;(x,\,f(x))\;$ ( because it belongs to the graph of $\;f\;$), but also as $\;(x,\,g(x))\;$ ( because it belongs to the graph of g as well), so we get that $\;f(x)=g(x)\;$ for some $\;x\;$ common to both functions' domain. That's all. In our case, $\;f(x)=3x^2+2kx+5\,,\,\,g(x)=2x-k\;$ ... Sep 21, 2019 at 21:12
• thx again, it makes more sense to me now Sep 21, 2019 at 21:54

Set the two equations equal to get $$2x - k = 3x^2 + 2kx + 5.$$ Now turn that into a single quadratic and solve. If the roots involve complex numbers (i.e. if $$b^2 - 4ac < 0$$), then there are no (real) intersections,