Which polynomial has zeros \\(\\sqrt{7}\\) and \\(-\\sqrt{7}\\)?

\\(x^2-7\\)

The sum of the zeros is \\(\sqrt{7}+(-\sqrt{7})=0\\) and the product is \\(\sqrt{7}\times(-\sqrt{7})=-7\\). For a monic quadratic the form is \\(x^2-(\text{sum})x+(\text{product})\\), so the required polynomial is \\(x^2-7\\). Why other options fail: option B has constant +7 (product +7) giving imaginary roots; option C has sum \\(2\sqrt{7}\\) and product +7 (a double positive root), not \\(\\pm\sqrt{7}\\); option D matches the product -7 but its sum is \\(-2\sqrt{7}\\), not 0, so its roots are different. Exam tip: compute sum and product of given roots and substitute into \\(x^2-(\text{sum})x+(\text{product})\\).

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