[ Quadratic equations. Vieta's Theorem ]

Symmetric polynomials $\quad]$

For polynomials $f(x)=x^{2}+a x+b$ and $g(y)=y^{2}+p y+q$ with roots $x_{1}, x_{2}$ and $y_{1}, y_{2}$ respectively, express through $a, b, p, q$ their resultant

$$
R(f, g)=\left(x_{1}-y_{1}\right)\left(x_{1}-y_{2}\right)\left(x_{2}-y_{1}\right)\left(x_{2}-y_{2}\right) .
$$
$\left(x_{1}-y_{1}\right)\left(x_{1}-y_{2}\right)\left(x_{2}-y_{1}\right)\left(x_{2}-y_{2}\right)=g\left(x_{1}\right) g\left(x_{2}\right)=\left(-a x_{1}-b+p x_{1}+q\right)\left(-a x_{2}-b+p x_{2}+q\right)=(p-a)^{2} x_{1} x_{2}+(p-a)(q-b) \left(x_{1}+x_{2}\right)+(q-b)^{2}=$

$=(p-a)^{2} b-(p-a)(q-b) a+(q-b)^{2}=(p-a)(p b-a b-q a+a b)+(q-b)^{2}=(p-a)(p b-a q)+(q-b)^{2}$.

## Answer

$(p-a)(p b-a q)+(q-b)^{2}$