Can a quadratic equation have irrational roots? By extension, can any equation have irrational roots? If not, why? If it can, how would you visualize it? (geometrically). I want to add that I am a high school school student. Don't want people wasting time writing answers that I can not understand

• What do you mean by 'visualize' it? – Simon S Nov 7 '14 at 21:58
• Consider $x^2 = 2$ – ChocolateBar Nov 7 '14 at 21:58
• consider $x=\pi$ – John Joy Nov 7 '14 at 22:01
• Do you know what an irrational number is? – Américo Tavares Nov 7 '14 at 22:03
• Yes, one that can not be expressed as a ratio of two integers? – user140161 Nov 7 '14 at 22:06

$x^2 -a = 0$ will give alot of irrational numbers. An interesting fact is that if a is a prime, the solution will allways be an irrational number. I am not sure how you can visualize it, but $\sqrt{2} = 1.41421...$ The number will continue forever and you will never be able to get "onto it", just very close. Don't know any other way to look at it. therefor $(1.41421...)^2 -2 = 0.00000...$
Sure $x^2 - 2$ has irrational roots.
• I don't understand. Neither root here of $\sqrt{2}$ or $-\sqrt{2}$ is infinite, but they are irrational. – Simon S Nov 7 '14 at 22:02
• What's an indefinite value? What ever they may be, $\sqrt{2}$ and $-\sqrt{2}$ are well defined. – Simon S Nov 7 '14 at 22:06