Question

Let $\Delta =\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|$ and the equation $p{x}^2+q{x}^2+rx+s=0$ has roots, $a,b,c$ where $a,b,c\in R^{+}$

If $\Delta =27$ and $a^2+b^2+c^2=3,$ then

• A. $3p+2q=0$
• B. $4p+3q=0$
• C. $3p+q=0$
• D. None of the above

Answer 1

The correct option is C $3p+q=0$

Multiplying $R_1,R_2,R_3$ by $a,b,c$ respectively, and then taking $a,b,c$ common from $C_1,C_2,$ and $C_3,$ we get
$\Delta =\left|\begin{array}{ccc}-bc& ab+ac& ac+ab\\ ab+bc& -ac& bc+ab\\ ac+bc& bc+ac& -ab\end{array}\right|$

Now, using $C_2\to C_2-C_1$ and $C_3\to C_3-C_1,$ and then taking $(ab+bc+ca)$ common from $C_2$ and $C_3,$ we get

$\Delta =\left|\begin{array}{ccc}-bc& 1& 1\\ ab+bc& -1& 0\\ ac+bc& 0& -1\end{array}\right|\times (ab+bc+ca{)}^2$

Now, applying $R_2\to R_2+R_1,$ we get
$\Delta =\left|\begin{array}{ccc}-bc& 1& 1\\ ab& 0& 1\\ ac+bc& 0& -1\end{array}\right|(ab+bc+ca{)}^2$

Expanding along $C_2,$ we get
$\Delta =(ab+bc+ca{)}^2[ac+bc+ab]$
$=(ab+bc+ca{)}^3$

If $\Delta =27,$ then $(ab+bc+ca{)}^3=27$
$\Rightarrow ab+bc+ca=3$

and $a^2+b^2+c^2=3,$
$\therefore (a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\Rightarrow a+b+c=\pm 3$
$\Rightarrow a+b+c=3$ (since all the roots are positive)
$\Rightarrow 3p+q=0$