# What is polynomial of order 2 that have the root x = 1+2`sqrt3` ?

## Expert Answers You need to determine the second degree polynomial `ax^2 + bx + c = 0` , whose root is `x_1 = 1 + 2sqrt 3` .

You need ot remember that if an equation has an irrational root, then it also has as a root the conjugate of the irrational root, thus, the polynomial whose root is `x_1 = 1 + 2sqrt 3` , then it allows as root its conjugate, `x_2 = 1 - 2sqrt 3` .

By Lagrange resolvents, the second degree polynomial may be considered as `x^2 - px + q = 0` , where `p = x_1 + x_2` and `q = x_1*x_2.`

Hence, you need to evaluate the sum `p` and the product `q` , such that:

`p = 1 + 2sqrt 3 + 1 - 2sqrt 3 => p = 2`

`q = (1 + 2sqrt 3)(1 - 2sqrt 3) => q = 1^2 - (2sqrt 3)^2`

`q = 1 - 12 = -11`

Replacing `2` for `p` and -`11` for `q` in equation yields:

`x^2 - px + q = x^2 - 2x - 11 = 0`

Hence, evaluating the second degree polynomial, under the given conditions, using Lagrange resolvents yields `x^2 - 2x - 11 = 0.`

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