Question

Let \[a,b,c\] be in arithmetic progression. Let the centroid of the triangle with vertices \[\left( {a,c} \right),\left( {2,b} \right)\] and \[\left( {a,b} \right)\] be \[\left( {\dfrac{{10}}{3},\dfrac{7}{3}} \right)\]. 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:
A. \[\dfrac{{71}}{{256}}\]
B. \[\dfrac{{ - 69}}{{256}}\]
C. \[\dfrac{{69}}{{256}}\]
D. \[\dfrac{{ - 71}}{{256}}\]

Answer 1

Hint: An arithmetic progression, also known as an arithmetic sequence, is a set of integers in which the difference between successive terms remains constant.
The roots of the quadratic equation are the values of the dependent variable where the value of the equation becomes zero.
Here, in this question, we will first establish a relation between the terms of the centroid of the triangle and the arithmetic progression, and then, with the help of the results, we will compare the sum and the product of the quadratic equation to get our final result.
The point inside a triangle where all the medians of the triangle meet are known as the Centroid of the triangle.

Image: Triangle with medians
Formula Used:
Centroid, \[C\left( {x,y} \right) \equiv \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\] where, \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are the coordinates of the vertices of the triangle.
Sum of the roots of the quadratic equation, \[\alpha + \beta = \dfrac{{ - b}}{a}\]
Product of the roots of the quadratic equation, \[\alpha \beta = \dfrac{c}{a}\]
Where, \[\alpha \] and \[\beta \] are the roots of the quadratic equation \[a{x^2} + bx + c = 0\].
Complete step-by-step answer:
As \[a,b,c\] be in arithmetic progression so, the common difference between the consecutive terms will be the same.
So, we can write it as-
\[
  b - a = c - b \\
   \Rightarrow b + b = c + a \\
   \Rightarrow 2b = a + c - - - - (i) \\
 \]

The centroid of the triangle is given as-
\[C\left( {x,y} \right) \equiv \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\] where, \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are the coordinates of the vertices of the triangle.
According to the question, the vertices of the triangle are \[\left( {a,c} \right),\left( {2,b} \right)\] and \[\left( {a,b} \right)\] while the centroid is \[\left( {\dfrac{{10}}{3},\dfrac{7}{3}} \right)\].
Substituting the values of the coordinates of the vertices and the centroid in the formula of the centroid of the triangle, we get
\[
  \dfrac{{10}}{3} = \dfrac{{a + 2 + a}}{3} \\
   \Rightarrow 2a + 2 = 10 \\
   \Rightarrow a + 1 = 5 \\
   \Rightarrow a = 4 \\
 \]
And,
\[
  \dfrac{7}{3} = \dfrac{{c + b + b}}{3} \\
   \Rightarrow 2b + c = 7 - - - - (ii) \\
 \]
From equation (i), we have \[2b = a + c\] so, substituting the same in the equation \[2b + c = 7\] we get,
\[
  2b + c = 7 \\
   \Rightarrow a + c + c = 7 \\
   \Rightarrow a + 2c = 7 \\
 \]
Substituting the value \[a\] as \[4\] and solving for \[c\], we get
\[
  4 + 2c = 7 \\
   \Rightarrow c = \dfrac{{7 - 4}}{2} \\
   \Rightarrow c = \dfrac{3}{2} \\
 \]
Now, substituting the value of \[c\] as \[\dfrac{3}{2}\] in equation (ii) to determine the value of \[b\], we get
\[
  2b + c = 7 \\
   \Rightarrow 2b + \dfrac{3}{2} = 7 \\
   \Rightarrow b = \dfrac{{14 - 3}}{4} \\
   \Rightarrow b = \dfrac{{11}}{4} \\
 \]

The given quadratic equation is \[a{x^2} + bx + 1 = 0\]. So, substituting the values of \[a,b,c\], we get
\[
  a{x^2} + bx + 1 = 0 \\
   \Rightarrow 4{x^2} + \dfrac{{11}}{4}x + 1 = 0 \\
   \Rightarrow 16{x^2} + 11x + 4 = 0 \\
 \]
It is given that \[\alpha ,\beta \] are the roots of the equation \[16{x^2} + 11x + 4 = 0\] and we need to determine the value of \[{\alpha ^2} + {\beta ^2} - \alpha \beta \].
Re-writing the term \[{\alpha ^2} + {\beta ^2} - \alpha \beta \], we get
\[
  {\alpha ^2} + {\beta ^2} - \alpha \beta = {\alpha ^2} + {\beta ^2} - \alpha \beta + 2\alpha \beta - 2\alpha \beta \\
   = \left( {{\alpha ^2} + {\beta ^2} + 2\alpha \beta } \right) - 3\alpha \beta \\
 \]
Using the identity, \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], we can write the above equation as-
\[{\alpha ^2} + {\beta ^2} - \alpha \beta = {\left( {\alpha + \beta } \right)^2} - 3\alpha \beta - - - - (iii)\]
From the quadratic equation, \[16{x^2} + 11x + 4 = 0\] we can write
Sum of the roots as, \[\alpha + \beta = \dfrac{{ - 11}}{{16}}\]
And, the product of the roots as, \[\alpha \beta = \dfrac{4}{{16}}\]
Substituting the values of the sum and the product of the roots in equation (iii), we get
 \[
  {\alpha ^2} + {\beta ^2} - \alpha \beta = {\left( {\alpha + \beta } \right)^2} - 3\alpha \beta \\
   = {\left( {\dfrac{{ - 11}}{{16}}} \right)^2} - 3\left( {\dfrac{4}{{16}}} \right) \\
   = \dfrac{{121}}{{256}} - \dfrac{3}{4} \\
 \]
Solving further, we get
\[
  {\alpha ^2} + {\beta ^2} - \alpha \beta = \dfrac{{121 - 192}}{{256}} \\
   = \dfrac{{ - 71}}{{256}} \\
 \]
Hence, option (d) is the correct answer.
Note: Some candidates make mistakes by opting to solve the quadratic equation either by splitting the middle term or by the determinant method but, here we need to evaluate the value of the expression termed in the roots of the quadratic equation.