# How do you find all additional roots given the roots 3-sqrt2 and 1+sqrt3?

Oct 3, 2016

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

#### Explanation:

If the coefficients of a polynomial equation are rational, complex

roots and irrational roots occur in conjugate pairs.

For examples.

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

$\pm \sqrt{3} , \pm \sqrt{5} \mathmr{and} 1$.

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

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 $\sqrt{a + \sqrt{b}}$, the additional roots will be

$\sqrt{a - \sqrt{b}} , - \sqrt{a + \sqrt{b}} \mathmr{and} - \sqrt{a - \sqrt{b}} .$ Here, altogether it

is a double couple $\pm \sqrt{a \pm \sqrt{b}}$..,