For which values of p and q will \(x^2+px+q\) have zeros \(4+\sqrt{7}\) and \(4-\sqrt{7}\)?

p = -8, q = 9

The sum of the roots is \((4+\sqrt{7})+(4-\sqrt{7})=8\) and their product is \((4+\sqrt{7})(4-\sqrt{7})=16-7=9\). For the monic quadratic \(x^2+px+q\), sum of roots = \(-p\) and product = \(q\). So \(-p=8\) gives \(p=-8\) and \(q=9\). Hence A is correct. Option B flips the sign of p, while C and D have the wrong product q. Exam tip: use sum = -p and product = q or expand \((x-\alpha)(x-\beta)\) to find coefficients quickly.

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