So i'm doing some problems from quadratic equations, this one in particular is supposed to be solved with vieta's formulas, I'm quite lost at what to do since there is a minus here, for example in solving $x_1^2+x_2^2$ for the same equation, its just a matter of adding $+2x_1x_2 -2x_1x_2$ to it and then inserting vieta's formulas in, but I have no idea what to do with -.
\begin{aligned} & x^2-3 x-10=0 \\ & x_1^2-x_2^2=? \end{aligned}
On
The two zeroes $ \ x_1 \ $ and $ \ x_2 \ $ imply that $ \ x_1^2 - 3 x_1 - 10 \ = \ 0 $ and $ \ x_2^2 - 3 x_2 -10 \ = \ 0 \ \ . \ $ Subtracting the second equation from the first gives us $ \ x_1^2 - x_2^2 \ - \ 3·(x_1 - x_2) \ = \ 0 \ \Rightarrow \ x_1^2 - x_2^2 \ = \ 3·(x_1 - x_2) \ \ .^{*} \ $ Since the difference of the two (real) zeroes of a quadratic polynomial is $ \ | \ x_1 - x_2 \ | \ = \ \frac{\sqrt{\Delta}}{a} \ \ , \ \ (a \ = \ 1 \ $ here) with $ \ \Delta \ $ being the discriminant, it is straightforward to calculate the quantity sought. Because we are taking a difference, the order of the zeroes matters, so what we will obtain is $ \ | \ x_1^2 - x_2^2 \ | \ \ . \ $ (This problem is simple to check since the quadratic polynomial is easily factored.)
$ ^{*} $ Notice that this is a "disguised" way of writing $ \ x_1^2 - x_2^2 \ = \ (x_1 + x_2)·(x_1 - x_2) \ \ . $
HINT…try using $$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$$