For the equation \(n^2-2pn+(p^2-7p)=0\), what condition on \(p\) is necessary for \(n\) to have two real and distinct roots?
Answer and explanation
Correct answer: \(p>0\)
Here, \(a=1\), \(b=-2p\), and \(c=p^2-7p\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2-7p)=28p\). Two real and distinct roots require \(D>0\), so \(28p>0\), which gives \(p>0\). At \(p=0\), \(D=0\) and the roots are equal, while for \(p<0\), the roots are not real. Exam tip: for distinct real roots, always check the condition \(D>0\).
Frequently asked questions
What is the correct answer to this question?
\(p>0\)
Why is this the correct answer?
Here, \(a=1\), \(b=-2p\), and \(c=p^2-7p\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2-7p)=28p\). Two real and distinct roots require \(D>0\), so \(28p>0\), which gives \(p>0\). At \(p=0\), \(D=0\) and the roots are equal, while for \(p<0\), the roots are not real. Exam tip: for distinct real roots, always check the condition \(D>0\).
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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