Complex Roots $\alpha$ and $\beta$ satisfy the equation $(x-\alpha)(x-\beta) = 0$ but not $(x+\alpha)(x+\beta) = 0$

Question

I would like to ask a question about Complex Roots in Further Mathematics.

I am new to the subject and one of the statements given in the book without further explanation is

If the roots of the equation are $\alpha$ and $\beta$, the equation is $(x-\alpha)(x-\beta) = 0$
Hence, $(x-\alpha)(x-\beta) = x^2 - \alpha x - \beta x + \alpha \beta = x^2 - (\alpha + \beta)x + \alpha \beta$

Now, I have a question. Why is the equation given as:

$$(x-\alpha)(x-\beta) = 0$$

and not as

$$(x+\alpha)(x+\beta) = 0$$

Is there a reason for this?

Answer 1

Inserting $\alpha$ into $(x-\alpha)(x-\beta)$ gives you $(\alpha-\alpha)(\alpha-\beta)$, which you can quickly confirm is equal to $0$. Same with inserting $\beta$.

On the other hand, inserting into $(x+\alpha)(x+\beta)$ doesn't give you zero (except in very special circumstances). So in order to make sure that $\alpha$ and $\beta$ are indeed roots, $-$ is the operation of choice.

Answer 2

The definition of a root is $P(r)=0$. So if $P(x):=(x-\alpha)(x-\beta)$, then $P(\alpha)=P(\beta)=0$. On the other hand, if $Q(x):=(x+\alpha)(x+\beta)$, then it's not necessarily true that $Q(\alpha)=0$.

Answer 3

$x-\alpha$ has the root $\alpha$ while $x+\alpha$ has the root $-\alpha$. Simple as that.