Let a, b, c be in arithmetic progression.

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 $$\left( {{{10} \over 3},{7 \over 3}} \right)$$. 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 :

Accepted Answer

2b = a + c

$${{2a + 2} \over 3} = {{10} \over 3}$$ and $${{2b + c} \over 3} = {7 \over 3}$$

a = 4,

$$\left\{ \begin{pmatrix} 2b + c = 7 \hfill \
  2b - c = 4 \hfill \ \end{pmatrix}  \right\}$$, solving

$$b = {{11} \over 4}$$

$$c = {3 \over 2}$$

$$ \therefore $$ Quadratic Equation is $$4{x^2} + {{11} \over 4}x + 1 = 0$$

$$ \therefore $$ The value of

$${\alpha ^2} + {\beta ^2} - \alpha \beta $$

= $${\alpha ^2} + {\beta ^2} + 2\alpha \beta - 3\alpha \beta  $$

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

$$= {{121} \over {256}} - {3 \over 4} =  - {{71} \over {256}}$$

Solutions · 1