JEE PYQ: Quadratic Equation Question 7
Question 7 - 2021 (24 Feb Shift 2)
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 $ax^2 + bx + 1 = 0$, then the value of $\alpha^2 + \beta^2 - \alpha\beta$ is:
(1) $\frac{71}{256}$
(2) $-\frac{69}{256}$
(3) $\frac{69}{256}$
(4) $-\frac{71}{256}$
Show Answer
Answer: (4)
Solution
$2b = a + c$
$\frac{2a+2}{3} = \frac{10}{3}$ and $\frac{2b+c}{3} = \frac{7}{3}$
$a = 4$, $2b + c = 7$ and $2b - c = 4$, solving:
$b = \frac{11}{4}$
$c = \frac{3}{2}$
$\therefore$ Quadratic Equation is $4x^2 + \frac{11}{4}x + 1 = 0$
$\therefore$ The value of $(\alpha + \beta)^2 - 3\alpha\beta = \frac{121}{256} - \frac{3}{4} = -\frac{71}{256}$
BETA
