I have the following example of the use of Vieta's formulas in my textbook:
Let's have a quadratic equation of the form
$x^2+x+q=0$
The following conditions apply: the equation has two roots, such as $x_1^2+x_2^2=1$. We need to find $q$.
The texbook says the following solution is possible:
$x_1^2+x_2^2 = x_1^2 + 2x_1x_2 + x_2^2 - 2x_1x_2 = (x_1 + x_2)^2 - 2x_1x_2$,
hence
$x_1^2+x_2^2 = 1 - 2q$,
hence
$1 - 2q = 1 \iff q=0$
The question I have is why did the authors decide that $(x_1 + x_2)^2=1$.
Because by Vieta's formulas $$x_1+x_2=-\frac 11=-1.$$
Note that in general, if $$ax^2+bx+c=0\ \ \ (a\not=0)$$ has roots $\alpha,\beta$, then the followings hold : $$\alpha+\beta=-\frac ba,\ \ \ \alpha\beta=\frac ca.$$