If \\(m+4)x^2-2(m+2)x+m=0\\) with \\(m\ne -4\\), what will be the nature of its roots?

Correct answer: Two real and distinct roots

Here, \(a=m+4\), \(b=-2(m+2)\), and \(c=m\). The condition \(m\ne-4\) ensures that \(a\ne0\), so the equation is genuinely quadratic. Its discriminant is \(\Delta=b^2-4ac=4(m+2)^2-4m(m+4)=16\). Since \(\Delta>0\), the roots are real and distinct. Equal roots would require \(\Delta=0\), so option B is incorrect. Exam tip: \(\Delta>0\), \(\Delta=0\), and \(\Delta<0\) indicate distinct real, equal real, and non-real roots, respectively.