- #1
SwimmingGoat
- 7
- 0
Problem:
[itex]q(x)=x^2-14\sqrt{2}x+87[/itex]. Find 4th degree polynomial [itex]p(x)[/itex] with integer coefficients whose roots include the roots of [itex]q(x)[/itex]. What are the other two roots of [itex]p(x)[/itex]?
I found that the two roots of [itex]q(x)[/itex] are [itex]x=7\sqrt{2}-\sqrt{11}[/itex] and [itex]x=7\sqrt{2}+\sqrt{11}[/itex]. Since they are conjugates of each other, I have no idea what to guess the other roots could be of my fourth-degree polynomial.
I started out with trying to get rid of the [itex]14\sqrt{2}[/itex] like so:
[tex](x^2-14\sqrt{2}+87)(x+14\sqrt{2})[/tex] but I ended up with
[tex](x^3+87x+1218\sqrt{2}-392)[/tex] Going to the fourth degree looked like a headache, and I felt I wasn't on the right track, so I stopped there.
Any ideas?
[itex]q(x)=x^2-14\sqrt{2}x+87[/itex]. Find 4th degree polynomial [itex]p(x)[/itex] with integer coefficients whose roots include the roots of [itex]q(x)[/itex]. What are the other two roots of [itex]p(x)[/itex]?
I found that the two roots of [itex]q(x)[/itex] are [itex]x=7\sqrt{2}-\sqrt{11}[/itex] and [itex]x=7\sqrt{2}+\sqrt{11}[/itex]. Since they are conjugates of each other, I have no idea what to guess the other roots could be of my fourth-degree polynomial.
I started out with trying to get rid of the [itex]14\sqrt{2}[/itex] like so:
[tex](x^2-14\sqrt{2}+87)(x+14\sqrt{2})[/tex] but I ended up with
[tex](x^3+87x+1218\sqrt{2}-392)[/tex] Going to the fourth degree looked like a headache, and I felt I wasn't on the right track, so I stopped there.
Any ideas?