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

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### 2 Answers

Given the one of the roots for the quadratic equation is: ( 2- 4sqrt3)

Then the other root is: (2+ 4sqrt3)

Since we have the roots we can form a quadratic equation by multiplying the factors:

Let the quadratic function be f(x):

==> f(x) = (x-x1) *(x-x2)

where x1 and x2 are the roots:

==> f(x) = ( x- (2-4sqrt3) ( x- (2+4sqrt3)

= x^2 -(2+4sqrt3)x +(2-4sqrt3)x + (2-4sqrt3)(2+4sqrt3)

= x^2 -2x - 4sqrt3 - 2x +4sqrt x4 - 44

= x^2 - 4x - 44

**==> f(x) = x^2 - 4x - 44**

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

Here we are given one root x1 = 2-4(3)^0.5.

Let the other root be x2.

Then we can form the quadratic equation like (x-x1)(x-x2) = 0. Or

x^2 - (x1+x2)x +(x1x2.

Therefore x1+x2= must be rational.

x1x2 = rational.

Therefore we chse the other x2 = 2+3(4)^(0.5) so that

x1+x2 = (2-4(3)^0.5) +(2+4(3)^0.5 = 4 is rational and

x1*x2 = (2-4(3)^0.5) *(2+4(3)^0.5 )= 2^2 - 4^2(3) = -44 is also rational .

Therefore therequired quadratic eqtian is :

x^2 - 4x - 44 = 0.