Question

If $\alpha$ and $\beta$ are the roots of $x^2-3px+p^2=0$ such that ${\alpha }^2+{\beta }^2=7/4$ then values of $p$ are

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

Answer 1

The correct option is D $1/2,-1/2$

$x^2-3px+p^2=0sum=\alpha +\beta =3pproduct=\alpha \beta =p^2{\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta \frac{7}{4}={\left(3p\right)}^2-{2p}^2\frac{7}{4}=7p^2{p}^2=\frac{1}{4}p=\pm \frac{1}{2}$