Let $\alpha \neq \beta$ be the roots of $x^2 + ax + b \in\Bbb R[x]$. Prove $\exists c \in \Bbb R$ such that $\alpha-\beta=c$ or $\alpha-\beta=ci$.

Question

Just a heads up: "$a$" and "$α$" are different

Let $a,b \in \Bbb R$ and suppose $a^2 − 4b \neq 0$. Let $\alpha$ and $\beta$ be the (distinct) roots of the polynomial $x^2 + ax + b$. Prove that there is a real number $c$ such that either $\alpha − \beta = c$ or $\alpha − \beta = ci$.

I have no idea how to prove this mathematically. Can someone explain how they would this, including how they would implement this using a proof tree?

This is what I was trying to do.

$$(x - \alpha)(x - \beta) = x^2 + ax + b$$

$$x^2 - \alpha x - \beta x + \alpha \beta = x^2 + ax + b$$

$$-\alpha x - \beta x = ax$$

$$-x(\alpha + \beta) = ax$$

$$(\alpha + \beta) = -a$$

$$\alpha \beta = b$$

However, I'm confused where to go from here and wondering if what I'm doing is wrong.

Answer 1

1. If $a^2 - 4 b > 0$ then the roots are : $$\alpha = \dfrac{-a + \sqrt{a^2 - 4 b}}{2} \text{ and } \beta = \dfrac{-a - \sqrt{a^2 - 4 b}}{2}$$ so : $$\alpha - \beta = \sqrt{a^2 - 4 b}$$ Let $c = \sqrt{a^2 - 4 b}$ then $c \in \mathbb{R}$ and $\alpha - \beta = c$.
2. If $a^2 - 4 b < 0$ then the roots are : $$\alpha = \dfrac{-a + i \sqrt{4b - a^2}}{2} \text{ and } \beta = \dfrac{-a - i \sqrt{4 b - a^2}}{2}$$ so : $$\alpha - \beta = i \sqrt{4 b - a^2}$$ Let $c = \sqrt{4 b - a^2}$ then $c \in \mathbb{R}$ and $\alpha - \beta = i c$.

Answer 2

Instead of factoring, you can apply the formula for the quadratic equation directly.

If $Ax^2 + Bx + C = 0$, then the two roots are given by

$$\frac{1}{2A} \left[-B \pm \sqrt{B^2 - 4AC}\right].$$

This means that (when the roots are distinct), the difference in the two roots will be

$$\frac{1}{2A} \times 2 \times \pm \sqrt{B^2 - 4AC} ~: ~B^2 - 4AC \neq 0.$$

In your problem, you have :

• $A = 1.$

• $B = a.$

• $C = b.$

So, the difference in the two roots is

$$\pm \sqrt{a^2 - 4b}. \tag1 $$

The expression in (1) above will either be real or imaginary, depending on whether $(a^2 - 4b)$ is positive or negative.

Answer 3

Solving the equation $x^2+ax+b=0$, we have $$x=\frac{-a\pm\sqrt{a^2-4b}}{2},$$ or more specifically, wlog, we can write $$ \alpha=\frac{-a+\sqrt{a^2-4b}}{2},\quad \beta=\frac{-a-\sqrt{a^2-4b}}{2}.$$ The difference, $\alpha-\beta$, can then be simplified: $$\alpha-\beta=\sqrt{a^2-4b}.$$

We are given $a^2-4b\neq0$, and this leads into two cases:
If $a^2-4b>0$, then $\alpha-\beta=c$ (the square root of a positive number is a real number).
If $a^2-4b<0$, then $\alpha-\beta=ci$ (the square root of a negative number is an imaginary number).

Answer 4

Here's an answer that needs very little algebra.

If the roots are real their difference $c$ is real.

If the roots are complex they are conjugates, so their difference is a real multiple of $i$.