Question

If $(\alpha ,\beta ),(\beta ,\gamma ),(\gamma ,\alpha )$ are respectively the roots of $x^2-2px+2=0;x^2-2qx+3=0;x^2-2rx+6=0$ where $\alpha ,\beta ,\gamma$ are all positive, then the value of $p+q+r$ is

• A. $1$
• B. $2$
• C. $0$
• D. $6$

Answer 1

The correct option is D $6$

$(\alpha ,\beta ),(\beta ,\alpha )$ and $(\gamma ,\alpha )$ are roots of $x^2-2px+2=0$, $x^2-2qx+3=0$ &$x^2-2rx+6=0$ respectively
So,
$\alpha +\beta =2p\to i$
$\beta +\gamma =2q\to ii$
$\gamma +\alpha =2r\to iii$
$\alpha \beta =2\to iv$
$\beta \gamma =3\to v$
$\gamma \alpha =6\to vi$
$iv\times v$
$=\alpha {\beta }^2\gamma =6$
${\beta }^2\left(6\right)=6$
${\beta }^2=1$
$\beta \pm 1$
$\gamma =\pm 3$
$\alpha =\pm 2$
$p+q+r=\frac{\alpha +\beta }{2}+\frac{\beta +\gamma }{2}+\frac{\gamma +\alpha }{2}$
From $i,ii,$ and $iii$
$=\frac{2(\alpha +\beta +\gamma )}{2}$
$=2+1+3$
$=6$