Question

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

Answer 1

\[2 x^2 + 2\left( p + q \right)x + p^2 + q^2 = 0\]

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)

\[\alpha\beta = \frac{p^2 + q^2}{2}\]

\[\left( \alpha - \beta \right)^2 = \left( \alpha + \beta \right)^2 - 4\alpha\beta\]

\[ = \left( p + q \right)^2 - 4\left( \frac{p^2 + q^2}{2} \right)\]

\[ = \left( p + q \right)^2 - 2\left( p^2 + q^2 \right)\]

\[ = - \left( p - q \right)^2 . . . . . \left( B \right)\]

\[A + B = \left( p + q \right)^2 - \left( p - q \right)^2 \]

\[ = \left( p + q - p + q \right)\left( p + q + p - q \right) \left[ \because x^2 - y^2 = \left( x - y \right)\left( x + y \right) \right]\]

\[ = 4pq\]

\[AB = \left( p + q \right)^2 \left[ - \left( p - q \right)^2 \right]\]

\[ = - \left( p^2 + q^2 + 2pq \right)\left( p^2 + q^2 - 2pq \right)\]

\[ = - \left[ \left( p^2 \right)^2 + \left( q^2 \right)^2 - 2 p^2 q^2 \right]\]

\[ = - \left( p^2 - q^2 \right)^2\]

The general form of a quadratic equation is
\[x^2 - \left( A + B \right)x + AB = 0\]

Putting the value of A and B we get \[x^2 - 4pqx - \left( p^2 - q^2 \right)^2 = 0\]