Form a quadratic equation with rational coefficients having 2 - 4(3)^.5 as one of its roots.

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( X - 2 + 4 (3)^0.5) ( x - 2 - 4(3)^0.5)

= X^2 - 4x + (2-4(3)^0.5) (2+4(3)^0.5

= X^2 - 4x + 4 - 16 (3)

= x^2 - 4x - 44

The quadrartic form should be (x+a)(x+b)

so, since 2- 4(3)^5 is one of the roots

(x-(2 - 4(3)^0.5)) * (x-(2+4(3)^0.5))

= x^2 -4x +4- 16*3

= x^2 -4x + 4-48

=x^2-4x-44 (quad. form) ax^2+bx+c=0

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