For real numbers \(p\) and \(q\), if \(q\ne0\), what is the nature of the roots of the equation \(x^2-2px+(p^2+q^2)=0\)?
Answer and explanation
Correct answer: No real roots
Here, \(a=1\), \(b=-2p\), and \(c=p^2+q^2\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2+q^2)=-4q^2\). Since \(q\ne0\), \(D<0\), so the equation has no real roots. In fact, its roots are \(p\pm iq\). Exam tip: For a quadratic equation, \(D<0\) immediately implies that the roots are not real.
Frequently asked questions
What is the correct answer to this question?
No real roots
Why is this the correct answer?
Here, \(a=1\), \(b=-2p\), and \(c=p^2+q^2\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2+q^2)=-4q^2\). Since \(q\ne0\), \(D<0\), so the equation has no real roots. In fact, its roots are \(p\pm iq\). Exam tip: For a quadratic equation, \(D<0\) immediately implies that the roots are not real.
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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