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