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