If the zeroes are \\( \\frac{3+\\sqrt{5}}{2} \\) and \\( \\frac{3-\\sqrt{5}}{2} \\), what is the monic quadratic polynomial?

\\(x^2-3x+1\\)

Let the roots be \\(
\\alpha=\\frac{3+\\sqrt{5}}{2},\\quad \\\beta=\\frac{3-\\sqrt{5}}{2}
\\).\n\nSum: \\(
\\alpha+\\beta=\\frac{3+\\sqrt{5}}{2}+\\frac{3-\\sqrt{5}}{2}=3
\\).\nProduct: \\(
\\alpha\\beta=\\frac{(3+\\sqrt{5})(3-\\sqrt{5})}{4}=\\frac{9-5}{4}=1
\\).\nA monic quadratic with these roots is \\(
x^2-(\text{sum})x+(\text{product})=x^2-3x+1
\\).\nThe closest distractor \\(x^2-3x-1\\) only has the constant term sign wrong, so it is incorrect. Exam tip: use the relation \\(x^2-(r_1+r_2)x+r_1r_2\\) immediately when roots are given.

This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.