Question

$\alpha$ and $\beta$ are the roots of the equation $x^2+px+p^3=0$ , $p\ne 0$ . If the point $(\alpha ,\beta )$ lies on the curve $x=y^2$, then the roots of the given equation are

• A. $4,-2$
• B. $4,2$
• C. $1,-1$
• D. $1,1$

Answer 1

The correct option is A $4,-2$

It is given that the point $(\alpha ,\beta )$ lies on the curve
$x=y^2$
Hence the roots are $({\beta }^2,\beta )$.
Hence sum of roots is
$=-p$
$={\beta }^2+\beta$.
${\beta }^2+\beta +p=0$ ...(i)
And
Product of roots is
$=p^3$
$={\beta }^3$
Hence
$p=\beta$.
Substituting in equation i.
$p^2+p+p=0$
$p^2+2p=0$
$p(p+2)=0$
$p=0$ and $p=-2$
Since $p\ne 0$.
Hence
$\beta =p=-2$
And
${\beta }^2=p^2=4$
Thus the roots are
${\beta }^2,\beta =(-2,4)$