The value of the parameter 'a' for which the quadratic equation $(1 - 2a){x^2} - 6ax - 1 = 0$ and $a{x^2} - x + 2 = 0$ have at least one root common , are

To find the common root, we can express x from the second equation as x=2-ax^2 or use cross-multiplication on the coefficients. Using the condition (c1*c2 - a1*a2)^2 = (a1*b2 - a2*b1)(b1*c2 - b2*c1), we substitute the coefficients a1=1-2a, b1=-6a, c1=-1, a2=a, b2=-1, and c2=2. This sets up the equation ((-1)(2) - (1-2a)(a))^2 = ((1-2a)(-1) - (a)(-6a))((-1)(2) - (-1)(-1)). Simplifying this gives (2a^2-a-2)^2 = (6a^2-2a-1)(-3), which reduces to the factored form (2a-1)(3a-1)(9a-2)=0. Solving this yields the possible values for a as 1/2, 1/3, and 2/9.

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