Question

If $\alpha$ and $\beta$ are the roots of $x^2+px+q=0$ from the equation whose roots are ${(\alpha -\beta )}^2$ and ${(\alpha +\beta )}^2$

Answer 1

$x^2+px+q=0$

$\alpha ,\beta$ are roots of above $e{q}^n$ , so

$\alpha +\beta =\frac{-p}{1}$

$\alpha \beta =q$

Since 2 roots of other $e{q}^n$ are : $(\alpha -\beta {)}^2$ & $(\alpha +\beta {)}^2$

so

sum of roots $\Rightarrow (\alpha -\beta {)}^2+(\alpha +\beta {)}^2$

$=2({\alpha }^2+{\beta }^2)-2\alpha \beta +2\alpha \beta$

$=2\left(\right(\alpha +\beta {)}^2-2\alpha \beta )$

$=2(p^2-2q)$

product of roots $\Rightarrow (\alpha -\beta {)}^2(\alpha +\beta {)}^2$

$\Rightarrow \left(\right(\alpha +\beta {)}^2-4\alpha \beta {)}^2(\alpha +\beta {)}^2$

$\Rightarrow (p^2-4q{)}^2(p^2)$

$\therefore$ The other $e{q}^n$ looks like :-

$x^2$ -(sum of roots)x+(product of roots ) = 0

$x^2-\left(2\right(p^2-2q\left)\right)x+(p^2-4q{)}^2(p{)}^2=0$

$x^2-(2p^2-4q)x+p^2\left(p^2\right(p^2-4q{)}^2=0$