The quadratic equation $a_1x^2-a_2x+a_3=0$ $a_1, a_2, a_3\in \mathbb N$ has two distinct roots $\in(1,2)$, then what is the least value of $a_1$

algebra-precalculusquadratics

If the quadratic equation $f(x)=a_1x^2-a_2x+a_3=0$  where $a_1, a_2, a_3\in \mathbb N^+$ has two distinct roots belonging to the interval $(1,2)$, then what is the least value of $a_1$?

my attempt

$D\gt0, f(1)\gt0, f(2)\gt0$ since natural number $a_1\ge1$ so leading coefficient is +ve i.e. an upward parabola.

$$\implies a_2^2-4a_1a_3\gt0$$

$$a_1-a_2+a_3\gt0$$

$$4a_1-2a_2+a_3\gt0$$

So Now I have three unknown and three inequalities how to handle it? What should I do now?

Edit 1

Since $f(1), f(2)\neq0$ but are natural number. So more precisely I can say the following

$$a_1-a_2+a_3\ge1$$

$$4a_1-2a_2+a_3\ge1$$

Edit 2

Let $\alpha, \beta$ be the roots of this quation then, $$\alpha+\beta=\dfrac {a_2}{a_1} \qquad \alpha \beta = \dfrac{a_3}{a_1}$$
 I get one more inequality

$${\alpha+\beta\over 2}= \dfrac{a_2}{2a_1}\in (1\; ,\; 2)$$

$$\implies2a_1<a_2<4a_1$$

$$\implies 4a_1-a_2>0 \\
a_2-2a_1>0$$

Now as $1\le a_1<\dfrac{a_2^2}{4a_3}\\
a_2^2>4a_1$ THEN I HAVE

$$a_1^2-a_1a_2+\dfrac{a_2^2}{4}>a_1$$

 $$4a_1^2-2a_1a_2+\dfrac{a_2^2}{4}>a_1$$

$$\implies (2a_1-a_2)^2>4a_1 \\
(4a_1-a_2)^2>4a_1$$

still no progress.

Best Answer

You're on the right track with the observations that $f(1) \geq 1, f(2) \geq 1$. How we can exploit this information?

Steps towards a solution. If you're stuck, explain what you've tried.

  1. Recall that $f(1) = a(1-\alpha) ( 1-\beta)$, and $ f(2) = a(2-\alpha)(2 - \beta)$.
  2. So, we have $ a^2(1-\alpha) ( 1-\beta)(2-\alpha)(2 - \beta) = f(1)f(2) \geq 1$.
  3. Show that $(\alpha - 1)(2- \alpha) \leq \frac{1}{4}$, with equality iff $ \alpha = \frac{3}{2}$.
  4. Hence, $(1-\alpha) ( 1-\beta)(2-\alpha)(2 - \beta) < \frac{1}{16}$.
    • Why do we have strict inequality?
  5. Thus $ \frac{ a^2}{16} > a^2 (1-\alpha) ( 1-\beta)(2-\alpha)(2 - \beta) \geq 1 $
    • So $ a \geq 5$.
  6. We're not yet done as we still need to show that $ a = 5$ can be achieved.
    • Consider $f(x) = 5x^2 + bx + c$.
    • Observe that since $ 2 > \frac{ 5^2}{16} > f(1)f(2)$. Thus we require $f(1) = 1, f(2) = 1$.
    • Solving this system gives us $ 5 + b + c = 1, 20 + 2b + c = 1 \Rightarrow b = -15, c = 11$.
    • A quick check of $ - \frac{b}{2a} \in (1,2) $ demonstrates that this quadratic has distinct roots between $(1, 2)$, hence we are done.

Notes

  • If we couldn't show strict inequality in step 4, then we'd only have $ a \geq 4$ in step 5. To continue from here, we could subsequently eliminate $ a= 4$ (EG by showing that no such quadratic exists per step 6).
  • In particular, $5x^2 - 15x + 11$ is the unique quadratic that satisfies the conditions with minimal $a$.
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