If the zeroes of (x^2+bx+12) are (2+\sqrt{7}) and (2-\sqrt{7}), what is the error?

Product is (-3), so constant term cannot be (12)

For a monic quadratic \(x^2+bx+12\), the product of its zeroes must equal the constant term, because the product is \(\frac{c}{a}\) and here \(a=1\). The proposed zeroes are \(2+\sqrt7\) and \(2-\sqrt7\). Their product is \((2+\sqrt7)(2-\sqrt7)=2^2-(\sqrt7)^2=4-7=-3\). Therefore the constant term should be \(-3\), not 12.

The zeroes are real because both expressions contain the real number \(\sqrt7\), and their sum is \(4\), which is rational. Thus options C and D are false. Option B correctly identifies the inconsistency: the product is \(-3\), so a constant term of 12 cannot be correct. The supplied answer and explanation are mathematically consistent.

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