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

Answer: (D)

None of the option is correct.
Since, a, b, c are in AP

$\therefore 2b=a+c$

Now, $a+2+a=10$

$\therefore a=4$

And, $c+2b=7$

$\Rightarrow c+a+c=7$

$\Rightarrow 2c=3$

$\therefore c=\frac{3}{2}$

$\therefore b=\frac{7-c}{2}=\frac{11}{4}$

Now,

$\alpha +\beta =\frac{-b}{a}=-\frac{11}{16}$

$\alpha \times \beta =\frac{c}{a}=\frac{3}{8}$

Now, ${\alpha }^2+{\beta }^2-\alpha \beta =(\alpha +\beta {)}^2-3\alpha \beta$

$\Rightarrow (\frac{-11}{16}{)}^2-3\frac{3}{8}=\frac{-167}{256}$