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 $\left(\frac{10}{3}, \frac{7}{3}\right)$. If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+1=0$, 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

Answer: (D)

$\frac{a+2+a}{3}=\frac{10}{3}$
$a=4$
and $\frac{c+b+b}{3}=\frac{7}{3}$
$c+2 b=7$
also $2 b=a+c$
$2 b-a+2 b=7$
$b=\frac{11}{4}$
now $4 x^{2}+\frac{11}{4} x+1=0 < ^{\alpha}_{\beta}$
$\alpha^{2}+\beta^{2}-\alpha \beta=(\alpha+\beta)^{2}-3 \alpha \beta$
$=\left(\frac{-11}{16}\right)^{2}-3\left(\frac{1}{4}\right)$
$=\frac{121}{256}-\frac{3}{4}=\frac{-71}{256}$