Question

Let $a,b,c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a,c),$ $(2,b)$ and $(a,b)$ be $(\frac{10}{3},\frac{7}{3}).$ If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+1=0,$ then the value of ${\alpha }^2+{\beta }^2-\alpha \beta$ is

• A. $\frac{71}{256}$
• B. $-\frac{69}{256}$
• C. $\frac{69}{256}$
• D. $-\frac{71}{256}$

Answer 1

The correct option is D $-\frac{71}{256}$

$2b=a+c$
$\frac{2a+2}{3}=\frac{10}{3}$ and $\frac{2b+c}{3}=\frac{7}{3}$
$\Rightarrow a=4$
$\begin{array}{c}2b+c=7\\ 2b-c=4\end{array}\}$, solving the two equations,
$b=\frac{11}{4}$ and $c=\frac{3}{2}$
$\therefore$ Quadratic equation is $4x^2+\frac{11}{4}x+1=0$
$\therefore$ The value of $(\alpha +\beta {)}^2-3\alpha \beta =\frac{121}{256}-\frac{3}{4}=-\frac{71}{256}$