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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 :

JEE · Math · previous-year question

  1. A.$${{69} \over {256}}$$
  2. B.$${{71} \over {256}}$$
  3. C.$$ - {{71} \over {256}}$$correct
  4. D.$$ - {{69} \over {256}}$$

Answer

C. $$ - {{71} \over {256}}$$

Explanation

2b = a + c $${{2a + 2} \over 3} = {{10} \over 3}$$ and $${{2b + c} \over 3} = {7 \over 3}$$ a = 4, $$\left\{ \begin{matrix} 2b + c = 7 \hfill \\ 2b - c = 4 \hfill \\\end{matrix} \right\}$$, 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 $${\alpha ^2} + {\beta ^2} - \alpha \beta $$ = $${\alpha ^2} + {\beta ^2} + 2\alpha \beta - 3\alpha \beta $$ = $${(\alpha + \beta )^2} - 3\alpha \beta $$ $$= {{121} \over {256}} - {3 \over 4} = - {{71} \over {256}}$$

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