Question 986181
Since {{{sqrt(4)=2}}}, we know that {{{2-sqrt(4)=2-2 = 0}}}. They made this root a bit more complicated than it had to be to confuse you. The three roots, as simple as they can get, are given as {{{5+sqrt(3)}}}, {{{-17}}}, and {{{0}}}



Rule: if you have a polynomial with rational coefficients, and you have a root {{{a+b*sqrt(c)}}}, then {{{a-b*sqrt(c)}}} must also be a root as well.


So if {{{5+sqrt(3)}}} is a root, then {{{5-sqrt(3)}}} must also be a root (note: a = 5, b = 1, c = 3 in this case)


Final Answer: <font color="red">Choice D)</font> {{{5-sqrt(3)}}}