Quadratic Equation Question 1

Question 1 - 24 January - Shift 1

Let $\alpha$ be a root of the equation $(a-c) x^{2}+(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix $ \begin{bmatrix} \alpha^{2} & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{bmatrix} $ is singular. Then the value of $\frac{(a-c)^{2}}{(b-a)(c-b)}+\frac{(b-a)^{2}}{(a-c)(c-b)}+\frac{(c-b)^{2}}{(a-c)(b-a)}$ is

(1) 6

(2) 3

(3) 9

(4) 12

Show Answer

Answer: (2)

Solution:

Formula: Roots under special case (vi)

$\Delta=0= \begin{vmatrix} \alpha^{2} & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{vmatrix} $

$\Rightarrow \alpha^{2}(c-b)-\alpha(c-a)+(b-a)=0$

It is singular when $\alpha=1$

$\frac{(a-c)^{2}}{(b-a)(c-b)}+\frac{(b-a)^{2}}{(a-c)(c-b)}+\frac{(c-b)^{2}}{(a-c)(b-a)}$

$ = \frac{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}{(a-b)(b-c)(c-a)}$

$=3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3$