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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?

questionasker
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1 Answers1

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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 $$a \in [-1.5, -1)$$

Rhys Hughes
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  • Actually, how do I tighten my range, as I don't know how to relate $a$ and $x$ in such a manner that I could use the domain. – questionasker Sep 24 '20 at 16:59
  • Added to my answer. Hope it helps – Rhys Hughes Sep 24 '20 at 17:28
  • Wait, but $-2$ also works. – questionasker Sep 24 '20 at 18:38
  • When $a=-2$, our solution pairs are $$(x,y)=(1+\phi^\pm, 2+\phi^\pm)$$ where $\phi^\pm=\frac{1\pm\sqrt5}{2}$. The positive versions are outside the range, as $\phi^+\approx 1.618 > 1$. Recall that our condition is that $x\leq 2$ and $y\leq 3$ – Rhys Hughes Sep 24 '20 at 19:15
  • Yeah, but graph $x^2 - 2x + 2$ and take a look at their intersections with the line segment – questionasker Sep 24 '20 at 19:26
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    I did.. The orange square in that graph represents the restrictions on $x,y$, and so you have a solution whenever the red and purple lines intersect twice inside that square. You can check other values by scrolling the $a$, but notice when $a=-2$, the second intersection (at $(1+\phi^+, 2+\phi^+)$) falls outside that square. – Rhys Hughes Sep 24 '20 at 19:31
  • Nevermind, I made a mistake in my graphing. Thanks! – questionasker Sep 24 '20 at 19:59