If \(r\) is any real number, what is the correct conclusion about the nature of the roots of the equation \(x^2-2rx+(r^2+9)=0\)?
Answer and explanation
Correct answer: No real roots
The discriminant of the quadratic is \(D=b^2-4ac=(-2r)^2-4(1)(r^2+9)=-36\). It remains negative for every real value of \(r\), so the equation has no real roots. In fact, its roots are \(r\pm3i\), which are complex and non-real. Exam tip: If \(D<0\), a quadratic equation 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 the quadratic is \(D=b^2-4ac=(-2r)^2-4(1)(r^2+9)=-36\). It remains negative for every real value of \(r\), so the equation has no real roots. In fact, its roots are \(r\pm3i\), which are complex and non-real. Exam tip: If \(D<0\), a quadratic equation 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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