If \(k\) is a real number and the roots of the equation \(x^2-2(k+1)x+(k^2+4)=0\) are real, what is the correct condition on \(k\)?
Answer and explanation
Correct answer: \(k\geq\frac{3}{2}\)
For a quadratic equation to have real roots, its discriminant must satisfy \(D\geq0\). Here, \(a=1\), \(b=-2(k+1)\), and \(c=k^2+4\), so \(D=b^2-4ac=4(k+1)^2-4(k^2+4)=8k-12\). Thus, \(8k-12\geq0\), giving \(k\geq\frac{3}{2}\). At \(k=\frac{3}{2}\), the roots are equal and real, so option C is too restrictive. Exam tip: use \(D\geq0\) for real roots and \(D>0\) only for distinct real roots.
Frequently asked questions
What is the correct answer to this question?
\(k\geq\frac{3}{2}\)
Why is this the correct answer?
For a quadratic equation to have real roots, its discriminant must satisfy \(D\geq0\). Here, \(a=1\), \(b=-2(k+1)\), and \(c=k^2+4\), so \(D=b^2-4ac=4(k+1)^2-4(k^2+4)=8k-12\). Thus, \(8k-12\geq0\), giving \(k\geq\frac{3}{2}\). At \(k=\frac{3}{2}\), the roots are equal and real, so option C is too restrictive. Exam tip: use \(D\geq0\) for real roots and \(D>0\) only for distinct real roots.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
