If the roots of the equation $\displaystyle ax^{2} + bx + c = 0$ are equal and $\displaystyle f\left ( x \right ) = \begin{vmatrix}ax+b & x^{2} & 4a \ bx+c & x & b \ cx+a & 1 & 4c\end{vmatrix}$, then the possible roots of $\displaystyle f\left ( x \right ) = 0$ is/are

Since the roots of ax^2 + bx + c = 0 are equal, the discriminant b^2 - 4ac equals zero, so b^2 = 4ac. Expanding the determinant f(x) gives a complex polynomial, but the lowest power of x in the expansion comes exclusively from the diagonal product (ax+b)(x)(4c). Multiplying this diagonal yields 4acx^2 + 4bcx, where every other product in the determinant expansion contributes terms with x^3, x^4, or x^5. Because b^2 = 4ac, the entire expanded polynomial has x factored out of every term. Therefore, f(x) = 0 always has x = 0 as one of its roots.

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