Question

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

• A. $3p+2q=0$
• B. $4p+3q=0$
• C. $3p+q=0$
• D. none of these

Answer 1

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

$\Delta =\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}-bc& b(b+c)& c(c+b)\\ a(a+c)& -ac& c(c+a)\\ a(a+b)& b(b+a)& -ab\end{array}\right|$
$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}-abc& ab(b+c)& ac(c+b)\\ ab(a+c)& -abc& bc(c+a)\\ ac(a+b)& bc(b+a)& -abc\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}-bc& a(b+c)& a(c+b)\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$R_1\to R_1+R_2+R_3$
$\Delta =\left|\begin{array}{ccc}ab+bc+ac& ab+bc+ac& ab+bc+ac\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$\Delta =(ab+bc+ac)\left|\begin{array}{ccc}1& 1& 1\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\Delta =(ab+bc+ac)\left|\begin{array}{ccc}0& 0& 1\\ ab+bc+ac& -(ab+bc+ac)& b(c+a)\\ 0& ab+bc+ac& -ab\end{array}\right|$
$\Delta =(ab+bc+ac{)}^2\left|\begin{array}{ccc}0& 0& 1\\ 1& -1& b(c+a)\\ 0& 1& -ab\end{array}\right|$
$\Delta =(ab+bc+ac{)}^2$ ....(i)
Since, a,b,c are the roots of the equation $p{x}^3+q{x}^2+rx+s=0$
$\Rightarrow ab+bc+ca=\frac{r}{p}$ and $abc=-\frac{s}{p},a+b+c=-\frac{q}{p}$
Now, given $\Delta =27$
$\Rightarrow (ab+bc+ac{)}^2=27$
$\Rightarrow ab+bc+ac=3\sqrt{3}$
$\Rightarrow \frac{r}{p}=3\sqrt{3}$
Also, $2(a^2+b^2+c^2)=(a+b+c{)}^2-2(ab+bc+ac)$
$\Rightarrow 4=\frac{q^2}{p^2}-6\sqrt{3}=p^{2}8\sqrt{3}=-q^2$
$3p+q=0$
Hence, option 'C' is correct.