Question

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation $2x^2+2\left(p+q\right)x+p^2+q^2=0$

Answer 1

$2x^2+2\left(p+q\right)x+p^2+q^2=0$
Let the roots of the given quadratic equation be $\alpha$ and $\beta$.
Sum of roots, $\alpha$ + $\beta$ = $\frac{-2\left(p+q\right)}{2}=-\left(p+q\right)$
$\Rightarrow {\left(\alpha +\beta \right)}^2={\left(p+q\right)}^2$ .....(A)
$\alpha \beta =\frac{p^2+q^2}{2}$
$$\begin{gathered} {\left(\alpha -\beta \right)}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta \\ ={\left(p+q\right)}^2-4\left(\frac{p^2+q^2}{2}\right) \\ ={\left(p+q\right)}^2-2\left(p^2+q^2\right) \\ =-{\left(p-q\right)}^2.....\left(B\right) \end{gathered}$$
$$\begin{gathered} A+B={\left(p+q\right)}^2-{\left(p-q\right)}^2 \\ =\left(p+q-p+q\right)\left(p+q+p-q\right)\left[\because x^2-y^2=\left(x-y\right)\left(x+y\right)\right] \\ =4pq \end{gathered}$$
$$\begin{gathered} AB={\left(p+q\right)}^2\left[-{\left(p-q\right)}^2\right] \\ =-\left(p^2+q^2+2pq\right)\left(p^2+q^2-2pq\right) \\ =-\left[{\left(p^2\right)}^2+{\left(q^2\right)}^2-2p^2{q}^2\right] \\ =-{\left(p^2-q^2\right)}^2 \end{gathered}$$
General form of quadratic equation is
$x^2-\left(A+B\right)x+AB=0$
Putting the value of A and B we get
$x^2-4pq-{\left(p^2-q^2\right)}^2=0$.