A number puzzle gives the equation \(n^2-2pn+(p^2-5p)=0\). What condition on \(p\) is necessary for the equation to have two real and distinct values of \(n\)?
Answer and explanation
Correct answer: \(p>0\)
Here, \(a=1\), \(b=-2p\), and \(c=p^2-5p\). Therefore, the discriminant is \(D=b^2-4ac=(-2p)^2-4(p^2-5p)=20p\). A quadratic equation has two real and distinct roots only when \(D>0\). Hence, \(20p>0\), which gives \(p>0\). For \(p=0\), the discriminant is zero and the roots are equal; for \(p<0\), the roots are non-real. Exam tip: For distinct real roots, always apply 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 \(c=p^2-5p\). Therefore, the discriminant is \(D=b^2-4ac=(-2p)^2-4(p^2-5p)=20p\). A quadratic equation has two real and distinct roots only when \(D>0\). Hence, \(20p>0\), which gives \(p>0\). For \(p=0\), the discriminant is zero and the roots are equal; for \(p<0\), the roots are non-real. Exam tip: For distinct real roots, always apply 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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