If \(p\) is a real constant, what is the nature of the roots of the equation \(x^2-2px+p^2=0\)?
Answer and explanation
Correct answer: Always real and equal
For this quadratic equation, \(a=1\), \(b=-2p\), and \(c=p^2\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-2p)^2-4(1)(p^2)=0\). Hence the roots are always real and equal. In fact, the equation can be written as \((x-p)^2=0\), so both roots are \(x=p\). Option B is incorrect because \(\Delta=0\) does not give distinct roots. Exam tip: \(\Delta=0\) indicates equal real roots.
Frequently asked questions
What is the correct answer to this question?
Always real and equal
Why is this the correct answer?
For this quadratic equation, \(a=1\), \(b=-2p\), and \(c=p^2\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-2p)^2-4(1)(p^2)=0\). Hence the roots are always real and equal. In fact, the equation can be written as \((x-p)^2=0\), so both roots are \(x=p\). Option B is incorrect because \(\Delta=0\) does not give distinct roots. Exam tip: \(\Delta=0\) indicates equal 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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