If the parabola $x^2 + ax + 2$ and the line segment connecting the points $(0,1)$ and $(2,3)$ (including the two endpoints) intersect 2 times at distinct points, find the range of $a.$
I've determined that $a = -2$ seems to be the only currently working number, but I'm not sure how to prove it. Can someone give me a hint?
It's quite simple to see that the line segment is $f: [0,2]\to [1,3]: f(x)=x+1$
We equate this to your curve: $$x^2+ax+2=x+1\implies x^2+(a-1)x+1=0$$
This has the solutions (by QF): $$x=\frac{(1-a)\pm\sqrt{(a-1)^2-4}}{2}=\frac{(1-a)\pm\sqrt{a^2-2a-3}}{2}\tag1$$ It has two unique, real solutions whenever the discriminant, $a^2-2a-3>0$. i.e. $a<-1, a>3$
Now you must take into consideration that the solutions must exist in the domain $[0,2]$. Use this to tighten your range.
When $a>3$, suppose $a=3+t$ for $t>0$, using $(1)$ we have $$x=\frac{-(t+2)\pm\sqrt{t(t+4)}}{2}<0$$ by AM-GM. (See if you can spot how)
When $a<-1$, again say $a=-(1+t), t>0$, we get $$x=\frac{(t+2)\pm\sqrt{t(t+4)}}{2}>0$$
for the same reason. Our solutions exist in this range, but after a certain point, where the second intersection is at the very end of the line segment, every other solution is too large.
Thus, we need to find the $a$ for which the second intersection meets at $(2,3)$, as anything beyond that will be out of our range. It's easy to see that $4+2a+2=3\implies a =-1.5$ and so our range is $$a \in [-1.5, -1)$$