#x^4-10x^2+1=0# has one root #x=sqrt(2)+sqrt(3)#. What are the other three roots and why?

1 Answer
Jun 14, 2015

The other three roots are #x = sqrt(2)-sqrt(3)#, #x = -sqrt(2)+sqrt(3)# and #x = -sqrt(2)-sqrt(3)#. As to the why, let me tell you a story...

Explanation:

Mr Rational lives in the town of Algebra.

He knows all the numbers of the form #m/n# where #m# and #n# are integers and #n != 0#.

He is quite happy solving polynomials like #3x+8=0# and #6x^2-5x-6=0#, but there are many that puzzle him.

Even an apparently simple polynomial like #x^2-2=0# seems insoluable.

His rich neighbour, Mr Real, takes pity on him. "What you need is what's called a square root of #2#. Here you go.". With these words, Mr Real hands over a mysterious shiny blue number called #R_2# to Mr Rational. All he is told about this number is that #R_2^2 = 2#.

Mr Rational goes back to his study and has a play with this mysterious #R_2#.

After a little while he finds that he can add, subtract, multiply and divide numbers of the form #a+b R_2# where #a# and #b# are rational and end up with numbers of the same form. He also notices that #x^2-2=0# has another solution, namely #-R_2#.

He is now able to solve not just #x^2-2=0#, but #x^2+2x-1=0# and many others.

Many other polynomials still evade solution. For example, #x^2-3=0#, but Mr Real is happy to give him a shiny green number called #R_3# that solves that one.

Mr Rational soon finds that he can express all the numbers he can make as #a+b R_2 + c R_3 + d R_2 R_3#, where #a#, #b#, #c# and #d# are rational.

One day Mr Rational has a go at solving #x^4-10x^2+1 = 0#. He finds that #x=R_2+R_3# is a solution.

Before he looks for more solutions, he bumps into his neighbour, Mr Real. He thanks Mr Real for the gift of #R_2# and #R_3#, but has a query about them. "I forgot to ask:", he says, "Are they positive or negative?". "I didn't think you'd care.", said Mr Real. "So long as you're solving polynomials with rational coefficients, it doesn't really matter. The rules you have found for adding, subtracting, multiplying and dividing your new numbers work just as well with either. Actually, I think the one you called #R_2# is what most people call #-sqrt(2)# and the one you called #R_3# is what most people call #sqrt(3)#".

So for Mr Rational's new numbers of the form #a+b R_2 + c R_3 + d R_2 R_3# it does not matter whether #R_2# and/or #R_3# are positive or negative from the point of view of solving polynomials with rational coefficients.