For the equation \(x^2-2rx+(r^2+25)=0\), where \(r\) is any real number, what is the correct conclusion about the nature of its roots?
Answer and explanation
Correct answer: No real roots
The discriminant of a quadratic equation is \(D=b^2-4ac\). Here, \(a=1\), \(b=-2r\), and \(c=r^2+25\), so \(D=(-2r)^2-4(1)(r^2+25)=-100\). Since \(D<0\) for every real value of \(r\), the equation has no real roots. Option B would apply only if \(D=0\), but the discriminant here is always negative. Exam tip: For a quadratic with real coefficients, \(D<0\) means that it has no real roots.
Frequently asked questions
What is the correct answer to this question?
No real roots
Why is this the correct answer?
The discriminant of a quadratic equation is \(D=b^2-4ac\). Here, \(a=1\), \(b=-2r\), and \(c=r^2+25\), so \(D=(-2r)^2-4(1)(r^2+25)=-100\). Since \(D<0\) for every real value of \(r\), the equation has no real roots. Option B would apply only if \(D=0\), but the discriminant here is always negative. Exam tip: For a quadratic with real coefficients, \(D<0\) means that it has no real roots.
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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