The values of the parameter $a$ for which the quadratic equations $(1 - 2a)x^{2} - 6ax - 1 = 0$ and $ax^{2} - x + 1 = 0$ have at least one root in common are
- $0, \dfrac{1}{2}$
- $\dfrac{1}{2}, \dfrac{2}{9}$
- $\dfrac{2}{9}$
- $0, \dfrac{1}{2}, \dfrac{2}{9}$
For common roots, the resultant of the two equations must be zero. Alternatively, subtracting the equations or using the condition for common roots leads to a = 2/9. Checking a=2/9 satisfies the condition.
Let alpha be the common root of the two quadratic equations. Using the condition for a common root, alpha^2 is equal to the determinant ( (-6a)(1) - (-1)(-1) ) divided by ( (1-2a)(1) - (-6a)(a) ), which simplifies to (-6a^2 - 1) divided by (1 - 2a + 6a^2). Substituting alpha and alpha^2 into the second equation ax^2 - x + 1 = 0 and replacing x with the corresponding cross-determinants ( (1-2a)(1) - a(-1) ) divided by ( (-6a)(a) - (-1)(1-2a) ), which equals 1 divided by (2a - 6a^2), yields a rational equation. Solving (a(-6a^2 - 1) - (1 - 2a + 6a^2) + (1 - 2a + 6a^2)) divided by (1 - 2a + 6a^2) equals 0 simplifies to -6a^3 - a = 0. Factoring out -a gives -a(6a^2 + 1) = 0. Since a cannot be zero for the second equation to remain quadratic, we check a = 0 and find it invalid. The only valid real solution is a = 2/9, but verifying the algebra shows a = 0 is required for consistency, while the sole acceptable nonzero parameter is 2/9.