Question

if $\alpha$, $\beta$ are the roots of the equation $x^2-px+q=0$ , then find the quadratic equation with the roots $({\alpha }^2-{\beta }^2)({\alpha }^3-{\beta }^3)$ and ${\alpha }^3{\beta }^2+{\alpha }^2{\beta }^3$

• A. $y^2+p\{q^2+(p^2-4q)(p^2-q)\}y+p^2{q}^2(p^2-4q)(p^2-q)=0$
• B. $y^2+p\{q^2+(p^2-4pq)(p^2-q)\}y+p^2{q}^2(p^2-4q)(p^2-q)=0$
• C. $y^2+p\{q^2+(p^2-4q)(p^2-q)\}y+4p^2{q}^2(p^2-4q)(p^2-q)=0$
• D. $y^2+p\{q^2+(p^2-4q^2)(p^2-q)\}y+p^2{q}^2(p^2-4q)(p^2-q)=0$

Answer 1

The correct option is A $y^2+p\{q^2+(p^2-4q)(p^2-q)\}y+p^2{q}^2(p^2-4q)(p^2-q)=0$

$x^2-px+q=0$
$\alpha +\beta =p$ and $\alpha \beta =q$

Now, $y_1=({\alpha }^2-{\beta }^2)({\alpha }^3-{\beta }^3)$

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

$=[{(\alpha +\beta )}^2-4\alpha \beta ][{(\alpha +\beta )}^2-\alpha \beta ](\alpha +\beta )=p(p^2-4q)(p^2-q)$

And $y_2={\alpha }^3{\beta }^2+{\alpha }^2{\beta }^3={\alpha }^2{\beta }^2(\alpha +\beta )=p{q}^2$

The quadratic equation is of the form: $y^2-(y_1+y_2)y+y_1{y}_2=0$
$\Rightarrow y^2+p\{q^2+(p^2-4q)(p^2-q)\}y+p^2{q}^2(p^2-4q)(p^2-q)=0$