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 $p{x}^3+q{x}^2+rx+s=0$ has roots $a,b,c$, where $a,b,c\in R^{+}$.The value of $\Delta$ is

• A. $\le 9r^2/p^2$
• B. $\ge 27s^2/p^2$
• C. $\le 27s^3/p^3$
• D. none of these

Answer 1

The correct option is B $\ge 27s^2/p^2$

$\Delta =\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b^2+ab& -ab\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}-bc& b(b+c)& c(c+b)\\ a(a+c)& -ac& c(c+a)\\ a(a+b)& b(b+a)& -ab\end{array}\right|$
$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}-abc& ab(b+c)& ac(c+b)\\ ab(a+c)& -abc& bc(c+a)\\ ac(a+b)& bc(b+a)& -abc\end{array}\right|$
$\Delta =\left|\begin{array}{ccc}-bc& a(b+c)& a(c+b)\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$R_1\to R_1+R_2+R_3$
$\Delta =\left|\begin{array}{ccc}ab+bc+ac& ab+bc+ac& ab+bc+ac\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$\Delta =(ab+bc+ac)\left|\begin{array}{ccc}1& 1& 1\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\Delta =(ab+bc+ac)\left|\begin{array}{ccc}0& 0& 1\\ ab+bc+ac& -(ab+bc+ac)& b(c+a)\\ 0& ab+bc+ac& -ab\end{array}\right|$
$\Delta =(ab+bc+ac{)}^2\left|\begin{array}{ccc}0& 0& 1\\ 1& -1& b(c+a)\\ 0& 1& -ab\end{array}\right|$
$\Delta =(ab+bc+ac{)}^2$ ....(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, $AM\ge GM$

$\Rightarrow \frac{ab+bc+ac}{3}\ge (a^2{b}^2{c}^2{)}^{1/3}$

$\Rightarrow ab+bc+ac\ge 3(abc{)}^{2/3}$

$\Rightarrow (ab+bc+ca{)}^3\ge 27\frac{s^2}{p^2}$

$\Rightarrow (ab+bc+ca{)}^2\ge 27\frac{s^2}{p^2}$

$\Rightarrow \Delta \ge 27\frac{s^2}{p^2}$

Hence, option 'B' is correct.