Let $a,b,c \in R$ such that no two of them are equal and satisfy $\left| {\begin{array}{*{20}{c}}{2a} & b & c\b & c & {2a}\c & {2a} & b\end{array}} \right| = 0$, then equation $24a{x^2} + 4bx + c = 0$ has
The determinant condition imposes a relationship among a, b, and c that forces the quadratic 24ax^2 + 4bx + c to change sign, or have a zero, within the interval [-1/2, 1/2]. Hence it must have at least one root in that interval. The other listed intervals are malformed or do not follow from the determinant condition.
Evaluating the given determinant results in the equation 2a^3 + b^3 + c^3 - 3abc = 0, which can be factored into (2a + b + c)((2a)^2 + b^2 + c^2 - 2ab - 2ac - bc) = 0. Since no two values among a, b, and c are equal, the second factor cannot be zero, leaving us with the condition 2a + b + c = 0. We evaluate the given quadratic polynomial 24ax^2 + 4bx + c at the endpoints of the interval [-1/2, 1/2]. At x = 1/2 the polynomial equals 6a + 2b + c, and at x = -1/2 it equals 6a - 2b + c. The sum of these two evaluations is 12a + 2c, which equals -4b after substituting b = -2a - c. The average of the evaluations is -2b, and adding the midpoint evaluation f(0) = c confirms the average is zero. Since the function value changes sign over this interval or averages to zero, Rolle's Mean Value Theorem guarantees that the equation has at least one root in the interval [-1/2, 1/2].