Question

Find the equation whose roots are ${(\alpha +\beta )}^{2}and{(\alpha -\beta )}^2,$ where $\alpha$ and $\beta$ are the roots of $2x^2+2(m+n)x+(m^2+n^2)=0.$

• A. $x^2-4mnx-{(m^2+n^2)}^2=0.$
• B. $x^2+2mnx-{(m^2-n^2)}^2=0.$
• C. $x^2-4mnx-{(m^2-n^2)}^2=0.$
• D. None of these

Answer 1

The correct option is C $x^2-4mnx-{(m^2-n^2)}^2=0.$

$\alpha +\beta =-(m+n),\alpha \beta =\frac{m^2+n^2}{2}$
$\therefore {(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta ={(m+n)}^2-2(m^2+n^2)=-(m-n{)}^2$
and ${(\alpha +\beta )}^2={(m+n)}^2.$
$\therefore S={(\alpha +\beta )}^2+{(\alpha -\beta )}^2={(m+n)}^2-{(m-n)}^2=4mn$
$P={(\alpha +\beta )}^2{(\alpha -\beta )}^2$
$=-{(m+n)}^2{(m-n)}^2=-{(m^2-n^2)}^2$
$\therefore$ Required equation is
$x^2-4mnx-{(m^2-n^2)}^2=0.$