If the zeroes of \(x^2+px+q\) are \(7+2\sqrt{3}\) and \(7-2\sqrt{3}\), what is the value of \(p+q\)?

23

For conjugate irrational roots the sum and product are easy: sum = \((7+2\sqrt{3})+(7-2\sqrt{3})=14\), product = \((7+2\sqrt{3})(7-2\sqrt{3})=49-(2\sqrt{3})^2=49-12=37\). For the monic quadratic \(x^2+px+q\), sum of roots = \(-p\) and product = \(q\). Hence \(p=-14\), \(q=37\) and \(p+q=37-14=23\). Closest distractor explanation: 37 is just the product \(q\), not \(p+q\); 49 would be wrong if one forgets to subtract \((2\sqrt{3})^2\). Exam tip: For a monic quadratic use sum = \(-p\), product = \(q\); compute \(p+q\) as \(q-\text{(sum of roots)}\) for speed.

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