If \(n\) is any real number, what is the nature of the roots of the equation \(x^2+2(n+2)x+n^2+4n+1=0\)?
Answer and explanation
Correct answer: Always real and distinct
For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=2(n+2)\), and \(c=n^2+4n+1\). Thus, \(D=4(n+2)^2-4(n^2+4n+1)=12\), which is positive for every real value of \(n\). Therefore, the roots are always real and distinct. Option B is incorrect because equal roots require \(D=0\). Exam tip: \(D>0\), \(D=0\), and \(D<0\) indicate real distinct, real equal, and non-real roots, respectively.
Frequently asked questions
What is the correct answer to this question?
Always real and distinct
Why is this the correct answer?
For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=2(n+2)\), and \(c=n^2+4n+1\). Thus, \(D=4(n+2)^2-4(n^2+4n+1)=12\), which is positive for every real value of \(n\). Therefore, the roots are always real and distinct. Option B is incorrect because equal roots require \(D=0\). Exam tip: \(D>0\), \(D=0\), and \(D<0\) indicate real distinct, real equal, and non-real roots, respectively.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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