How do you find all additional roots given the roots $3 - \sqrt{2}$ $3-sqrt2$ and $1 + \sqrt{3}$ $1+sqrt3$ ?

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Answer 1 · 2016-10-03T15:56:31

$3 + \sqrt{2} and 1 - \sqrt{3}$ $3+sqrt 2 and 1 - sqrt 3$

If the coefficients of a polynomial equation are rational, complex

roots and irrational roots occur in conjugate pairs.

For examples.

the roots of $x^{3} - x^{4} - 8 x^{3} + 8 x^{2} + 15 x - 15 = 0$ $x^3-x^4-8x^3+8x^2+15x-15=0$ are

$\pm \sqrt{3}, \pm \sqrt{5} and 1$ $+-sqrt 3, +-sqrt5 and 1$ .

The roots of $x^{3} + x^{2} + x + 1 = 0$ $x^3+x^2+x+1=0$ are $\pm i and - 1$ $+-i and -1$ .

Here the conjugates 3+sqrt 2 and 1 - sqrt 3 should also be the roots

of such an equation.

In the case of surd like $\sqrt{a + \sqrt{b}}$ $sqrt(a+sqrtb)$ , the additional roots will be

$\sqrt{a - \sqrt{b}}, - \sqrt{a + \sqrt{b}} and - \sqrt{a - \sqrt{b}} .$ $sqrt(a-sqrtb), -sqrt(a+sqrtb) and -sqrt(a-sqrtb).$ Here, altogether it

is a double couple $\pm \sqrt{a \pm \sqrt{b}}$ $+-sqrt(a+-sqrtb)$ ..,

How do you find all additional roots given the roots 3-sqrt23−√23-sqrt2 and 1+sqrt31+√31+sqrt3?