Quadratic Equations
Roots of quadratic equation and determinant condition
nta_pyq_2023_jan
Grade 11
Question:
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 $\dfrac{(a-c)^2}{(b-a)(c-b)} + \dfrac{(b-a)^2}{(a-c)(c-b)} + \dfrac{(c-b)^2}{(a-c)(b-a)}$ is
Step-by-Step Solution
Key Concept: If the matrix is singular, its determinant is zero. Expanding along the first row gives a condition on $\alpha$. Using $\alpha = 1$ (since $x=1$ is always a root of $(a-c)x^2+(b-a)x+(c-b)=0$) and algebraic identity $(a-b)^3+(b-c)^3+(c-a)^3 = 3(a-b)(b-c)(c-a)$.
Setting $\Delta = 0$ gives $\alpha^2(c-b) - \alpha(c-a) + (b-a) = 0$. Since $x=1$ satisfies $(a-c)+(b-a)+(c-b)=0$, $\alpha = 1$. The expression equals $\dfrac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} = \dfrac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} = 3$.
Correct Answer: 2