Question

Let $a,b$ and $c$ satisfy the system of equations $a+2b+3c=6,4a+5b+6c=12$ and $6a+9b=4$ . If the roots of the equation $\left(a+b+c\right)x^2-abcx+\left(a^{-1}+b^{-1}+c^{-1}\right)=0$ are $\alpha$ and $\beta$ , then $\frac{1}{\alpha }+\frac{1}{\beta }$ is equal to

• A. $243$
• B. $100$
• C. $\frac{243}{12}$
• D. $\frac{100}{243}$

Answer 1

Answer: (D)

Using cramer’s rule
$\Delta =\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 6& 9& 0\end{array}\right|=1\left(-54\right)-2\left(-36\right)+3\left(6\right)$
$=-54+72+18=36$
${\left(\Delta \right)}_1=\left|\begin{array}{ccc}6& 2& 3\\ 12& 5& 6\\ 4& 9& 0\end{array}\right|=6\left(-54\right)-2\left(-24\right)+3\left(88\right)$
$=-324+48+264=-12$
${\left(\Delta \right)}_2=\left|\begin{array}{ccc}1& 6& 3\\ 4& 12& 6\\ 6& 4& 0\end{array}\right|=1\left(-24\right)-6\left(-36\right)+3\left(-56\right)$
$=-24+216-168=24$
${\left(\Delta \right)}_3=\left|\begin{array}{ccc}1& 2& 6\\ 4& 5& 12\\ 6& 9& 4\end{array}\right|=1\left(-88\right)-2\left(-56\right)+6\left(6\right)$
$=-88+112+36=60$
$\Rightarrow a=\frac{{\Delta }_1}{\Delta }=-\frac{1}{3},b=\frac{{\Delta }_2}{\Delta }=\frac{2}{3},c=\frac{{\Delta }_3}{\Delta }=\frac{5}{3}$
$\Rightarrow$ equation is $2x^2+\frac{10}{27}x-\frac{9}{10}=0$
$\Rightarrow 540x^2+100x-243=0$
$\alpha +\beta =-\frac{5}{27},\alpha \beta =-\frac{9}{20}$
$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-5/27}{-9/20}=\frac{100}{243}$