constraints on the sum and product of roots of quadratic equation assuming less than unity roots

Question

I am solving a math contest problem. Assume we have the quadratic equation $x^2+a_1x+a_2=0$ where $a_1,a_2\in \mathbb{R}$ are real numbers. The roots of this equation can be found as (from equation it can be inferred that $a_1 = -(p_1+p_2), a_2 = p_1p_2$)

$$p_1,p_2 = -\frac{a_1}{2}\pm \sqrt{\frac{a_1^2}{4}-a_2}$$

Now assume that $|p_1|< 1, |p_2|< 1$, in other words the module of the roots is lesser than unity (in the case of real roots, it translates to the fact that their absolute value is less than one). The claim is twofold

1. $|a_2| < 1$

2. $|a_1| < a_2+1$

proving the first one very easy since $|a_2|=|p_1p_2| \le |p_1||p_2|< 1$. However, the second one is almost impossible to prove! If we assume the roots are real, then by using $|p_i|< 1$ we can say

$$-1< -\frac{a_1}{2}\pm \sqrt{\frac{a_1^2}{4}-a_2}< 1 \Rightarrow -1+\frac{a_1}{2} < \pm \sqrt{\frac{a_1^2}{4}-a_2}< 1 + \frac{a_1}{2}$$ then this gives two inequalities which can be manipulated to get similar results with a lot of mental gymnastics! But I don't know what to do in general! Is there any simpler way to reach the conclusion? What can be done in general case (complex conjugate roots and real distinct roots)?

Thank you!

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For the complex conjugate pair, we can write $p_1 = p_r+ip_i, p_2 = p_1-ip_i$ and by substitution we get two true statements

$$(1+p_r)^2+p_i^2 > 0 , (p_r-1)^2+p_i^2 > 0$$

and this prove the complex case. However, I am trying to find a way to drive the second statement using the given assumptions! Is there a way?

Answer 1

By Vieta, $p_1+p_2=-a_1$ and $p_1p_2=a_2$

As you say, it's trivial to show that $\big|a_2\big|<1$.

Let's assume that $\big|a_1\big| \ge a_2+1$ and try to reach a contradiction. Substituting and squaring both sides, \begin{align}\left(p_1+p_2\right)^2 &\ge \left(p_1 p_2+1\right)^2 \\ p_1^2+2p_1 p_2 +p_2^2 &\ge p_1^2 p_2^2+2p_1 p_2+1 \\ p_1^2 +p_2^2 &\ge p_1^2 p_2^2+1\\ 0&\ge\left(p_1^2-1\right)\left(p_2^2-1\right) \end{align}

If $p_1$ and $p_2$ are real, this means either $p_1^2-1\le 0$ and $p_2^2-1\ge 0$, or vice-versa. But this immediately contradicts the condition that both $\big|p_1\big|<1$ and $\big|p_2\big|<1$.

So the only possibility is that $p_1$ and $p_2$ are a complex conjugate pair. But then $p_1^2-1$ and $p_2^2-1$ are also a conjugate pair, whose product must be positive; contradiction!

Therefore $\big|a_1\big| < a_2+1$.

Answer 2

Following my original answer, Chris Lewis left a comment/question asking where I had considered the complex root case. My original answer did not consider this case; this was my oversight.

To make the answer easier to read, I have added an Addendum, which explores the complex root case. So, the entire first part of my answer, before the Addendum, assumes that both roots are real.

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The second assertion is equivalent to being given $~|p_1|, ~|p_2| < 1,~$ and being asked to prove that

$$|p_1 + p_2| < (p_1 p_2) + 1. \tag1 $$

Clearly, without loss of generality, neither of $~p_1,~p_2~$ is equal to zero.

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$\underline{\text{Case 1:} ~p_1, p_2 ~\text{have the same sign}}$
Here, because of the nature of the assertion in (1) above, you can assume, without loss of generality that $~p_1, ~p_2 ~$ are both positive.

So, you need to prove that

$$p_1 + p_2 < (p_1 p_2) + 1. \tag2 $$

You can assume that there exist $~a,b \in (0,1)~$ such that
$p_1 = 1 - a, ~p_2 = 1 - b \implies $
$p_1 + p_2 = 2 - (a+b),~$ and
$(p_1 p_2) + 1 = [(1 - a)(1 - b)] + 1 = 2 - (a+b) + (ab).$

So, within Case 1, the assertion is reduced to proving that

$$2 - (a+b) < 2 - (a+b) + (ab)$$

which is immediate.

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$\underline{\text{Case 2:} ~p_1, p_2 ~\text{have different signs}}$
Here, you can assume, without loss of generality, that $~-1 < p_1 < 0 < p_2 < 1.$

So, you need to prove that

$$|p_2 + p_1| < (p_1 p_2) + 1. \tag3 $$

Either $~|p_2| \geq |p_1|~$ or $~|p_2| < |p_1|.$

Assuming that $~|p_2| \geq |p_1|,~$ and letting $~q_1~$ denote $~-p_1,~$the assertion in (3) above is equivalent to proving that

$$p_2 - q_1 < 1 - (q_1 p_2).$$

You can assume that there exist $~a,b \in (0,1)~$ such that
$q_1 = 1 - a, ~p_2 = 1 - b \implies $

• Since $~q_1 \leq p_2, ~0 < b \leq a < 1.$

• $|p_2 + p_1| = p_2 - q_1 = a - b.$

• $(p_1 p_2) + 1 = [(a - 1)(1 - b)] + 1 = (a + b) - (ab)$

So, within Case 2, under the assumption that $~|p_2| \geq |p_1|,~$

the assertion is reduced to proving that

$$a - b < (a + b) - (ab) \iff ab < 2b$$

which is immediate.

So, again letting $~q_1 = - p_1,~$ the entire problem has been reduced to proving the assertion, under the assumption that $~0 < p_2 < q_1 < 1.$

Again, you can assume that there exist $~a,b \in (0,1)~$ such that
$q_1 = 1 - a, ~p_2 = 1 - b \implies $

• Since $~p_2 < q_1, ~0 < a < b < 1.$

• $|p_2 + p_1| = q_1 - p_2 = b - a.$

• $(p_1 p_2) + 1 = [(a - 1)(1 - b)] + 1 = (a + b) - (ab)$

So, in this specific situation, you are being asked to prove that

$$(b - a) < (a + b) - (ab) \iff ab < 2a$$

which is immediate.

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$\underline{\text{Addendum : exploring the complex root case}}$

Here, you have that

$$\frac{(a_1)^2}{4} < a_2. \tag 4 $$

and you are being asked to prove that

$$|a_1| < a_2 + 1. \tag 5$$

You are given that $~|p_1|, ~|p_2| < 1, ~$

and that $~p_2~$ is equal to one of the two values

$$\frac{-a_1}{2} \pm \sqrt{\frac{(a_1)^2}{4} - a_2}$$

$$=\frac{-a_1}{2} \pm i\sqrt{a_2 - \frac{(a_1)^2}{4}}. \tag6 $$

Since $~|p_2|^2 < 1,~$ you can conclude from (6) above that

$$\frac{(a_1)^2}{4} + \left[ ~a_2 - \frac{(a_1)^2}{4} ~\right] < 1 \implies a_2 < 1. \tag7 $$

So, combining the conclusions of (4) above with (7) above, you have that

$$0 \leq \frac{(a_1)^2}{4} < a_2 < 1 \tag8 $$

and you are being asked to prove that

$$|a_1| < a_2 + 1. \tag9 $$

Since both sides of (9) above are non-negative, this is equivalent to proving that

$$(a_1)^2 < (a_2)^2 + 2a_2 + 1,$$

where you know, from (8) above that

$$(a_1)^2 < 4a_2.$$

So, it is sufficient to prove that

$$4a_2 < (a_2)^2 + 2a_2 + 1 \iff 2a_2 < (a_2)^2 + 1. \tag{10}$$

The assertion on the right side of (10) above is immediate because:

• From (8) above, $~0 < a_2 < 1.$

• Therefore, $~0 < (1 - a_2)^2 = 1 + (a_2)^2 - 2a_2.$

Answer 3

I was pondering about this question the other day and got a better look at the problem and found out the intuition is nothing more than the cosine theorem! Let's assume $p_1 = re^{-j\theta}, p_1 = re^{j\theta}$. Therefore $a_1 = 2r\cos \theta, a_2 = r^2$. The inequality becomes

$$2r|\cos \theta|<r^2+1$$

If we look at the location of the root in the following figure

The cosine theorem indicates that $\overline{Ap_1}^2 = r^2 + 1 - 2r\cos \theta > 0$ and that gives the result. As long as $|p_1|<1, |p_2|<1$, this result will still be correct in the case of two distinct real roots as @Chris Lewis's answer shows above!

I also asked the designer of the question (one of our teachers) what the intuition behind the given relations is, and he did not give an answer, but he said this is a well-known result in digital signal processing, known as "stability triangle"!