The theorems I want to prove are:
$$\text {if } a + b + c = 0 \text{ then } x_1 = 1 \text { and } x_2= c/a$$ $$ \text{if } a - b + c = 0 \text{ then } x_1 = - 1 \text{ and } x_2 = -c/a $$
where $x_1$ and $x_2$ are the roots of the equation and $a,b$ and $c$ are the parameters. In my textbook the only thing which is said is that this is easily proven by Viet's formulas. However I can see how to prove it.
Let $P(x)=ax^2+bx+c$.
If $a+b+c=0$ this implies $P(1)=0$. This implies that $1$ is a root of $P(x)$. Note that the multiple of the two roots of a quadratic is $\frac{a}{c}$. However, one root is $1$. Thus, the other remaining solution is $\frac{a}{c}$.
In the same method, $a-b+c=0$ implies $P(-1)=0$, and so $-\frac{a}{c}$ is a solution.
In general, if $ak^2+bk+c=0$, one solution is $k$ and the other is $\frac{a}{ck}$, which one can prove using the method above.
