(1) the value of the parameter 'a' for which the quadratic equation (1-2a)x²-6ax-1=0 and ax²-x+1=0 have at least one root common are.

ans = 2/9

well simply I'll apply kamchor method..

in first and second equation , if they will be getting a common root that has a term containing sq.root in it, then we can equate their respective determinants but that gives no relevant result , so there is no term containing sq.root !! Thus both of the equations must simultaneously yield determinants that are perfect square .. If for second equation u try to do so , the other eqn. 's determinant doesnt become perfect square.

But to do so in first eqn. we make       determinant = b² - 4ac = 36a² - 8a +4    a perfect square by only   making  36a² = 8a,, it implies   a=2/9  

it gives atleast one common root for both equation as=3.   

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