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

1 Answer
Oct 3, 2016

Answer:

#3+sqrt 2 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^4-8x^3+8x^2+15x-15=0# are

#+-sqrt 3, +-sqrt5 and 1#.

The roots of #x^3+x^2+x+1=0# are #+-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+sqrtb)#, the additional roots will be

#sqrt(a-sqrtb), -sqrt(a+sqrtb) and -sqrt(a-sqrtb).# Here, altogether it

is a double couple #+-sqrt(a+-sqrtb)#..,