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 $a{x}^2+bx+1=0$, 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

The correct option is A

$\frac{71}{256}$

Explanation for the correct option:

Finding the value of ${\alpha }^2+{\beta }^2-\alpha \beta$

Given that,

$a,b,c$ be in arithmetic progression

therefore, $2b=a+c....\left(i\right)$

Since the centroid of the triangle with vertices $(a,c),(2,b)$ and $(a,b)$ be $\left(\frac{10}{3},\frac{7}{3}\right)$

Therefore,

$\left(\frac{a+2+a}{3},\frac{c+b+b}{3}\right)=\left(\frac{10}{3},\frac{7}{3}\right)$

$$\begin{gathered} \Rightarrow \frac{(2a+2)}{3}=\frac{10}{3} \\ \Rightarrow a=4..\left(ii\right) \end{gathered}$$

And,

$$\begin{gathered} \Rightarrow \frac{(2b+c)}{3}=\frac{7}{3} \\ \Rightarrow 2b+c=7...\left(iii\right) \end{gathered}$$

From $\left(i\right)\&\left(ii\right)$ we have'

$2b-c=4..\left(iv\right)$

Solving $\left(iii\right)\&\left(iv\right)$

$b=\frac{11}{4}\&c=\frac{3}{2}$

Substituting values of $a,b,c$ into the given quadratic equation

$\Rightarrow 4x^2+\frac{11}{4}x+1=0$

Since, $\alpha ,\beta$ are roots of this quadratic equation

So, sum of roots,

$$\begin{gathered} \alpha +\beta =-\frac{b}{a} \\ =-\frac{{\displaystyle \frac{11}{4}}}{4} \\ =-\frac{11}{16} \end{gathered}$$

And product of roots,

$$\begin{gathered} \alpha \beta =\frac{c}{a} \\ =\frac{{\displaystyle \frac{3}{2}}}{4} \\ =\frac{3}{8} \end{gathered}$$

Therefore, the value of the given expression

$$\begin{gathered} {(\alpha +\beta )}^2–3\alpha \beta ={\left(-\frac{11}{16}\right)}^2-3\left(\frac{3}{8}\right) \\ =-\frac{71}{256} \end{gathered}$$

Hence, the correct option is (A)