Hints to solve two quadratic polynomial.

Question

If $a,b$ are roots of the $x^{2}+px+1=0$ and $c,d$ are roots of the $x^{2}+qx+1=0$ then $E=(a-c)(b-c)(a+d)(b+d)=?$. Can we get this terms of $p,q$ ?

Answer 1

Apply Vedic Theorem in one variable quadratic equation, $x_1, x_2$ are roots of $ax^2+bx+c=0$, we imply that $x_1+x_2=-\frac{b}{a}$, $x_1 \cdot x_2 = \frac{c}{a}$. So we can get $a+b=-p$, $a \cdot b = 1$, $c+d=-q$, $c \cdot d = 1$. $$\begin{align*} E & = (ab -ac -bc + c^2)(ab + ad + bd + d^2) \\ & =^{a+b=-p} (1 + pc + c^2)(1 - pd + d^2) \\ & = 1 -pd + d^2 + pc -p^2cd +pcd^2 + c^2 -pc^2d +c^2d^2\\ & =^{cd=1,c^2d=c,cd^2=d} 1 - pd + d^2 + pc - p^2 + pd + c^2 -pc + 1\\ & = 1 - pd + pd + pc -pc + d^2 + c^2 -p^2 + 1\\ & = 2 + d^2+c^2 - p^2 \\ & = 2 + (d+c)^2 -2dc - p^2\\ & = 2 + q^2 -2 -p^2\\ & = q^2 -p^2\\ \end{align*}$$

To avoid too long expansion, in the second step, we can use $c^2+1 = -qc$, $d^2+1 = -qd$ to substitute them. Then we get $$\begin{align*} E & = (1+pc+c^2)(1-pd+d^2) \\ & = (pc-qc)(-pd-qd) \\ & = (p-q)c \cdot -(p+q)d\\ & = (p-q) \cdot -(p+q) \cdot cd\\ & = q^2- p^2\\ \end{align*}$$

This way is much simpler.

Answer 2

By Viete's formulas we have $a+b=-p$ and $ab=1$. Similarly $c+d=-q$ and $cd=1$. From this it follows for example that $c+\frac{1}{c}=d+\frac{1}{d}=-q$ and then: \begin{align}\frac{(a-c)(b-c)}{c}&=\frac{1+pc+c^2}{c}=c+\frac{1}{c}+p=p-q\\ \frac{(a+d)(b+d)}{d}&=\frac{1-pd+d^2}{d}=d+\frac{1}{d}-p=-p-q \end{align} Multiplying these together and using the fact that $cd=1$ we obtain $(p-q)(-p-q)=q^2-p^2$.

Answer 3

From the factorization of the first polynomial, and from Vieta's formulas

$$E=P(c)P(-d)\\ =(Q(c)+(p-q)c)(Q(d)-(p+q)d)\\ =-(p^2-q^2)cd\\ =q^2-p^2.$$