SOLUTION: Three roots of a fourth-degree polynomial equation with rational coefficients are 5+√3, –17, and 2-√4. Which number also is a root of the equation? A. 17 B. 2+√

Algebra ->  Rational-functions -> SOLUTION: Three roots of a fourth-degree polynomial equation with rational coefficients are 5+√3, –17, and 2-√4. Which number also is a root of the equation? A. 17 B. 2+√      Log On


   

Question 986181: Three roots of a fourth-degree polynomial equation with rational coefficients are 5+√3, –17, and 2-√4. Which number also is a root of the equation?
A. 17
B. 2+√4
C. 4-√2
D. 5-√3
How does this make sense? I thought that if you switched the addition/subtraction symbol, then you will have the other root. So shouldn't A, B, and D all be correct?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since sqrt%284%29=2, we know that 2-sqrt%284%29=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%2Bsqrt%283%29, -17, and 0


Rule: if you have a polynomial with rational coefficients, and you have a root a%2Bb%2Asqrt%28c%29, then a-b%2Asqrt%28c%29 must also be a root as well.

So if 5%2Bsqrt%283%29 is a root, then 5-sqrt%283%29 must also be a root (note: a = 5, b = 1, c = 3 in this case)

Final Answer: Choice D) 5-sqrt%283%29